Disclaimer: This is an old incomplete answer, now aged multiple years. I started this before I had any formal math education and I never came back to finish this.
To make our lives easier, we can separate the pieces by 3 types:
Edges (3 stickers)
Sides (2 stickers)
Centers (1 sticker)
Also note that im observing a cube with Yellow and Black( I call it White later on) colors for the top and bottom sides as my default color scheme.
(Other pieces follow Red, Blue, Orange, Green)
For the $3\times3\times3$ Rubik's cube, first we can solve for the middle parts.
We can conclude that if the color scheme is known, that all obscured centers can be determined if we have at least 2 neighbour ones.
This means we can peel off 4 centers if we keep 2 neighbour ones. This can be checked by playing with its own cube or simply by following the laws of legal positions and using something like a solver help you out. (This is because centers can't be really switched, but the pieces around them can so that it looks that way)
Then lets see how many stickers from the edge pieces can be obscured or peeled off, since any configuration of $n\times n\times n$ always has exactly 8 edges.
We can list them by the colors they contain:
(Red, Blue, Green, Orange, Yellow, White)
$$ R,G,Y $$
$$ R,G,W $$
$$ R,B,Y $$
$$ R,B,W $$
$$ O,G,Y $$
$$ O,G,W $$
$$ O,B,Y $$
$$ O,B,W $$
If you want to know how many stickers can be peeled off and still be able to identify each corner, you can ask yourself; How many stickers you can take so that when putting them back, you have only one possible way to do so?
I have found out the maximum number of 12 out of 24 stickers; by taking all R or O, then all B or G and finally all Y or W.
Here is an example:
$$ -,-,- $$
$$ -,-,W $$
$$ -,B,- $$
$$ -,B,W $$
$$ O,-,- $$
$$ O,-,W $$
$$ O,B,- $$
$$ O,B,W $$
Now by shuflling the order and rotation of all 8 edges in every way possible, there will be only one way to stick the removed stickers, thus you still can identify the pieces.
Now the Side pieces, which might be a bit tricky.
On the $3\times3\times3$ Rubik's cube, there are 12 Edge pieces:
$$ G,R $$
$$ R,B $$
$$ B,O $$
$$ O,G $$
$$ Y,G $$
$$ G,W $$
$$ W,B $$
$$ B,Y $$
$$ R,W $$
$$ W,O $$
$$ O,Y $$
$$ Y,R $$
Now you can try to do the same thing.
Firstly I tried to remove all White stickers, then it allows me to take one color from $Y,?$ pieces other than yellow, and after that nothing more can be taken without providing multiple solutions for the edges; so that was 5 stickers.
Then after other failed attempts, I found you can remove 6 stickers; one of each color so that there aren't multiple stickers of the same color standing without their second color, and I'm kinda sure it can't be done with more than 6 here.
If someone can do better here, please let me know.
So if I haven't made a mistake, for the $3\times3\times3$ Rubik's cube you can remove total of $12+6+4 = 22$ out of 54 stickers top (following the things I pointed out) without losing any information about the cube's states.
The $2\times2\times2$ Rubik's cube is made of only 8 edge pieces, so 12 out of 24 stickers can be removed here.
We can now try to generalize this to other $n\times n\times n$ cubes.
We now know that we can for;
$n=2$ take 12 out of 24
$n=3$ take 22 out of 54
For any $n\times n\times n$ cube, we always have 8 edges, so thats $+12$ obscured stickers.
For the side pieces, we have $n$ of each piece (sorted set of pieces):
$$ W,R $$
$$ W,B $$
$$ W,G $$
$$ W,O $$
$$ Y,R $$
$$ Y,B $$
$$ Y,G $$
$$ Y,O $$
$$ R,B $$
$$ R,G $$
$$ O,B $$
$$ O,G $$
There is nothing more to do here than apply the same thing I did in $3\times 3\times 3$ cube;
Remove 6 stickers, one of each color so that there aren't multiple stickers of the same color standing without their second color and do that for each new set of the side pieces.
This provides us with $6\times(n-2)$ pieces more to obscure.
Again, if you can do better with the sorted set I provided, please let me know.
So far then, the number of stickers we can obscure is: $$12+6\times(n-2)+C$$
Where $C$ is the number of "mobile" and "immobile" centers we can obscure, that appear after $n>3$ and have yet to be figured out.
(when $n=3$ then $c=4$ )
So now, the center pieces at $4\times 4\times 4$ cube and every other with $2k$ sides ($k>1$) are different than every $2k+1$ sided cubes;
The $2k+1$ like our standart $3\times3\times3$ cube have 6 "real centers" which are immobile and $4$ out of $6$ can be obscured, the rest one-sticker centers here are "mobile" and behave differently when being rotated.
Same goes for all of centers which are all "mobile" in $2k$ cubes.
How many of these mobile centers can be obscured? I would say for a starting bound, all of one color which is then:
$C = (n-2)^2$ for $n>3$
And gives us finally:
$$12+6\times(n-2)+(n-2)^2$$
Pieces we can surely obscure.
I haven't yet started checked if more can be obscured on these centers, but thats because its now end of the day and I don't have any more time, and now decided to post my progress so far.
I think you could take it from here to finish up the generalization and try to see if it is possible to obscure more than $(n-2)^2$ stickers using the color scheme and legal permutations, so maybe separate formulas for $2k$ and $2k+1$ can be found.
Update
Actually I don't think the mobile centers should provide us with any additional problems, thus we can take for $C$ that it is: $ = (n-2)^2 \times 4$ Since we need only 2 full center colors that will tell us the rest, most easily after solving the cube to its solved state.
Then we have: $$12+6\times(n-2)+4\times(n-2)^2$$
I have classes to be on early tomorrow morning now, so good night.
Edit: This should be computed and checked to make sure I haven't made
a mistake somewhere.
Best Answer
Let $\Pi$ be the space of colorings and fix an element $\pi\in\Pi$. There is a set-theoretic isomorphism given by $S_{54}\to\Pi:\sigma\mapsto\sigma(\pi)$. Thus we can consider the rubik's cube action as taking place on the symmetric group. In particular, we have an embedding $R\hookrightarrow S_{54}$ when we view $\pi$ as the standard configuration of the cube. The action of $R$ on $S_{54}$ is given by left multiplication.
When a subgroup $H\le G$ acts on $G$ by left multiplication, the orbits are of the form $Hg$, so we actually are looking at the right coset space $G/H$. The number of orbits is then $|G/H|$.
Therefore the number of colorings modulo rubics cube actions is
$$\frac{54!}{43252003274489856000}=5337179318013082492954222792315582522933248000000000.$$
I would personally avoid Burnside's lemma for this type of problem, where a direct computation is perfectly feasible while the group's structure (and therefore its fixed point sets) are more difficult to characterize than the original coset space's size.