Perl 6, 76 65 63 59 bytes

-4 bytes thanks to Phil H

221222==[~] (.skip Z-$_)X%12o*>>.&13*.ord+>3+?/#/-?/b/

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Explanation

*>>.& ... # Map notes to integers
 13*.ord # 13 * ASCII code: A=845 B=858 C=871 D=884 E=897 F=910 G=923
 +>3 # Right shift by 3: A=105 B=107 C=108 D=110 E=112 F=113 G=115
 # Subtracting 105 would yield A=0 B=2 C=3 D=5 E=7 F=8 G=10
 # but isn't necessary because we only need differences
 +?/#/ # Add 1 for '#'
 -?/b/ # Subtract 1 for 'b'

 o # Compose with block
 (.skip Z-$_) # Pairwise difference
 X%12 # modulo 12
 [~] # Join
 221222== # Equals 221222

Node.js v10.9.0, 78 76 71 69 bytes

a=>!a.some(n=>(a-(a=~([x,y]=Buffer(n),x/.6)-~y%61)+48)%12-2+!i--,i=3)

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How?

Each note $n$ is converted to a negative number in $[-118,-71]$ with:

[x, y] = Buffer(n) // split n into two ASCII codes x and y
~(x / .6) // base value, using the ASCII code of the 1st character
- ~y % 61 // +36 if the 2nd character is a '#' (ASCII code 35)
 // +38 if the 2nd character is a 'b' (ASCII code 98)
 // +1 if the 2nd character is undefined

Which gives:

 n | x | x / 0.6 | ~(x / 0.6) | -~y % 61 | sum
------+----+---------+------------+----------+------
 "Ab" | 65 | 108.333 | -109 | 38 | -71
 "A" | 65 | 108.333 | -109 | 1 | -108
 "A#" | 65 | 108.333 | -109 | 36 | -73
 "Bb" | 66 | 110.000 | -111 | 38 | -73
 "B" | 66 | 110.000 | -111 | 1 | -110
 "B#" | 66 | 110.000 | -111 | 36 | -75
 "Cb" | 67 | 111.667 | -112 | 38 | -74
 "C" | 67 | 111.667 | -112 | 1 | -111
 "C#" | 67 | 111.667 | -112 | 36 | -76
 "Db" | 68 | 113.333 | -114 | 38 | -76
 "D" | 68 | 113.333 | -114 | 1 | -113
 "D#" | 68 | 113.333 | -114 | 36 | -78
 "Eb" | 69 | 115.000 | -116 | 38 | -78
 "E" | 69 | 115.000 | -116 | 1 | -115
 "E#" | 69 | 115.000 | -116 | 36 | -80
 "Fb" | 70 | 116.667 | -117 | 38 | -79
 "F" | 70 | 116.667 | -117 | 1 | -116
 "F#" | 70 | 116.667 | -117 | 36 | -81
 "Gb" | 71 | 118.333 | -119 | 38 | -81
 "G" | 71 | 118.333 | -119 | 1 | -118
 "G#" | 71 | 118.333 | -119 | 36 | -83

We compute the pairwise differences modulo $12$ between these values.

The lowest possible difference between 2 notes is $-47$, so it's enough to add $4times12=48$ before applying the modulo to make sure that we get a positive result.

Because we apply a modulo $12$, the offset produced by a '#' is actually $36 bmod 12 = 0$ semitone, while the offset produced by a 'b' is $38 bmod 12 = 2$ semitones.

We are reusing the input variable $a$ to store the previous value, so the first iteration just generates $textNaN$.

For a major scale, we should get $[ textNaN, 2, 2, 1, 2, 2, 2 ]$.

We use the counter $i$ to compare the 4th value with $1$ rather than $2$.

Dart, 198 197 196 189 bytes

f(l)var i=0,j='',k,n=l.map((m)k=m.runes.first*2-130;k-=k>3?k>9?2:1:0;return m.length<2?k:m[1]=='#'?k+1:m[1]=='b'?k-1:k;).toList();for(;++i<7;j+='$(n[i]-n[i-1])%12');return'221222'==j;

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Loose port of the old Perl 6 answer https://codegolf.stackexchange.com/a/175522/64722

f(l)
 var i=0,j='',k,
 n=l.map((m)
 k=m.runes.first*2-130;
 k-=k>3?k>9?2:1:0;
 return m.length<2?k:m[1]=='#'?k+1:m[1]=='b'?k-1:k;
 ).toList();
 for(;++i<7;j+='$(n[i]-n[i-1])%12');
 return'221222'==j;

• -1 byte by using ternary operators for #/b


• -1 byte by using ifs instead of ternaries for the scale shifts


• -7 bytes thanks to @Kevin Cruijssen

Old version :

Dart, 210 bytes

f(l)var i=0,k=0,n='C':0,'D':2,'E':4,'F':5,'G':7,'A':9,'B':11,'b':-1,'#':1,j='',y=[0,0];for(;++i<7;j+='$(y[0]-y[1])%12')for(k=0;k<2;k++)y[k]=n[l[i-k][0]]+(l[i-k].length>1?n[l[i-k][1]]:0);return'221222'==j;

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Ungolfed:

f(l)
 var i=0,k=0,n='C':0,'D':2,'E':4,'F':5,'G':7,'A':9,'B':11,'b':-1,'#':1,j='',y=[0,0];
 for(;++i<7;j+='$(y[0]-y[1])%12')
 for(k=0;k<2;k++)
 y[k]=n[l[i-k][0]]+(l[i-k].length>1?n[l[i-k][1]]:0);

 return'221222'==j;

A whole step is 2, a quarter is 1. Mod 12 in case you jump to a higher octave.
Iterates through all notes and computes the difference between the ith note and the i-1th note.
Concatenates the result and should expect 221222 (2 whole, 1 half, 3 wholes).

• -2 bytes by not assigning 0 to k


• -4 bytes by using j as a String and not a List


• -6 bytes thanks to @Kevin Cruijssen by removing unnecessary clutter in loops

JavaScript (Node.js), 150 131 125 bytes

l=>(l=l.map(x=>'C0D0EF0G0A0B'.search(x[0])+(x[1]=='#'|-(x[1]=='b')))).slice(1).map((n,i)=>(b=n-l[i])<0?2:b)+""=='2,2,1,2,2,2'

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-19 bytes thanks to Luis felipe
-6 bytes thanks to Shaggy

Ungolfed:

function isMajor(l) 
 // Get tone index of each entry
 let array = l.map(function (x) -(x[1]=='b');
 );
 // Calculate deltas
 let deltas = array.slice(1).map(function (n,i) 
 // If delta is negative, replace it with 2
 // This accounts for octaves
 if (n - array[i] < 0) return 2;
 // Otherwise return the delta
 return n - array[i];
 );
 // Pseudo array-comparison
 return deltas+"" == '2,2,1,2,2,2';

C (gcc), -DA=a[i] + 183 = 191 bytes

f(int*a)char s[9],b[9],h=0,i=0,j=0,d;for(;A;A==35?b[i-h++-1]++:A^98?(b[i-h]=A*13>>3):b[i-h++-1]--,i++);for(;j<7;d=(b[j]-b[j-1])%12,d=d<0?d+12:d,s[j++-1]=d+48);a=!strcmp(s,"221222");

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Based on the Perl answer.

Takes input as a wide string.

Ungolfed:

int f(int *a)
	char s[9], b[9];
	int h, i, j;
	h = 0;
 for(i = 0; a[i] != NULL; i++)
		if(a[i] == '#')
			b[i-h-1] += 1;
			h++;
		
		else if(a[i] == 'b')
			b[i-1-h] -= 1;
			h++;
		
		else
			b[i-h] = (a[i] * 13) >> 3;
		
	
	for(j = 1; j < 7; j++)
		int d = (b[j] - b[j-1]) % 12;
		d = d < 0? d + 12: d;
		s[j-1] = d + '0';
	
	return strcmp(s, "221222") == 0;

Python 3, 175 136 134 bytes

def f(t):r=[ord(x[0])//.6%12+'b #'.index(x[1:]or' ')-1for x in t];return[12*(r[i+1]<r[i])+r[i+1]-r[i]for i in range(6)]==[2,2,1,2,2,2]

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An one-liner Python 3 implementation.

Thanks to @Arnauld for idea of calculate tones using division and modulo.

Thanks to @Jo King to save -39 bytes!

-2 bytes by removing parentheses.

[Python] 269 202 bytes

Improvements from Jo King:

p=lambda z:"A BC D EF G".index(z[0])+"b #".index(z[1:]or' ')-1
def d(i,j):f=abs(p(i)-p(j));return min(f,12-f)
q=input().replace(' ','').split(',')
print([d(q[i],q[i+1])for i in range(6)]==[2,2,1,2,2,2])

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Ungolfed, with test driver:

tone = "A BC D EF G" # tones in "piano" layout
adj = "b #" # accidentals

def note_pos(note):
 if len(note) == 1:
 note += ' '
 n,a = note
 return tone.index(n) + adj[a]

def note_diff(i, j):
 x, y = note_pos(i), note_pos(j)
 diff = abs(x-y)
 return min(diff, 12-diff)

def is_scale(str):
 seq = str.replace(' ','').split(',')
 div = [note_diff(seq[i], seq[i+1]) for i in (0,1,2,3,4,5)]
 return div == [2,2,1,2,2,2]

case = [
("C, D, E, F, G, A, B", True),
("C#, D#, E#, F#, G#, A#, B#", True),
("Db, Eb, F, Gb, Ab, Bb, C", True),
("D, E, Gb, G, A, Cb, C#", True),
("Eb, E#, G, G#, Bb, B#, D", True),

("C, D#, E, F, G, A, B", False),
("Db, Eb, F, Gb, Ab, B, C", False),
("G#, E, F, A, B, D#, C", False),
("C#, C#, E#, F#, G#, A#, B#", False),
("Eb, E#, Gb, G#, Bb, B#, D", False),
]

for test, result in case:
 print(test + ' '*(30-len(test)), result, 't',
 "valid" if is_scale(test) == result else "ERROR")

Ruby, 109 bytes

->s(0..11).any?s.map(w="Cef;DXg<E=Fhi>G j8A d9B:")[(w.index(""<<n.sum%107)/2-t)%12]*''=='CfDX<=h'

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