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.
The maximum orders for a $n\times n\times n$ Rubik's cube group:
\begin{align*}n=2: &\quad 3^2\cdot 5 = 45\\
n=3: &\quad 2^2\cdot 3^2\cdot 5\cdot 7 = 1260\\
n=4: &\quad 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17 = 765765\\
n=5: &\quad 2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 23 = 281801520\\
n\ge 6: &\quad 2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23 = 5354228880\end{align*}
In general, there are five types of piece to consider:
- For odd $n$, each face has a center. By convention, we treat these as fixed. If we don't, they can be rotated like a solid cube, for a copy of $S_4$. Highest order $4$, least common multiple $2^2\cdot 3=12$. As the largest possible order is divisible by $12$ for all odd $n\ge 3$, whether we include these doesn't affect the answer.
- For any $n\ge 2$, there are $8$ corner pieces which can be permuted and rotated, a semidirect product of $S_8$ and $(\mathbb{Z}/3)^8$ - except that that the "sum" of those rotations is zero, dropping one factor of $3$ from the latter part. The order of an element is at worst $3$ times the order of an element of $S_8$. Highest possible order $15\cdot 3=45$, least common multiple $2^3\cdot 3^2\cdot 5\cdot 7$.
- For any odd $n\ge 3$, there are $12$ centers of edges, which can be permuted and reflected. The number of reflections must also be even, for a semidirect product of $S_{12}$ and $(\mathbb{Z}/2)^{11}$. Highest possible order $60\cdot 2=120$, least common multiple $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11$.
- For any $n\ge 4$, there are $\left\lfloor\frac{n}{2}\right\rfloor-1$ orbits of 24 off-center edges, which can't change how far from the center of their edge they are. The pair of edges on each side are distinguishable due to being reflections of each other, and the group for each set of these pieces is a copy of $S_{24}$. Highest possible order $8\cdot 7\cdot 5\cdot 3 = 840$, least common multiple $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23$.
- For any $n\ge 4$, there are $\left\lfloor\frac{(n-2)^2}{4}\right\rfloor$ orbits of 24 off-center face pieces each. These can be rotated in their face, but not reflected; mirror images across a symmetry line within a face (in size $\ge 6$) are separate orbits. Each piece of the same color in one of these sets is indistinguishable, so that gets us an action of $S_{24}$, with $\frac{24!}{24^6}$ possible states. The possible orders are the same as in the previous case. In terms of what we need to solve the cube, there are three variants based on whether they lie on a symmetry line of a face - but they're all the same for the count we care about here.
So then, the highest possible order in any Rubik group is the least common multiple of all of these, namely $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23$. This is achieved for all $n\ge 6$; we have at least six of the size-24 orbits, which we can use to produce orders of $16\cdot 5$, $9\cdot 11$, $7\cdot 13$, $17$, $19$, and $23$.
Now, all of this assumes that we can modify the pieces in each orbit independently. That isn't quite true - several combinations must be even permutations. First, for odd $n$, the combined permutation of corners and edge centers is even; they're modified by face rotations, each of which is a 4-cycle of edges and a 4-cycle of corners. Second, for $n\ge 4$, the combined permutation of the corners and an orbit of pieces on face diagonals is even. Next, for $n\ge 6$, each mirrored pair of orbits of off-center face pieces must have a combined even permutation; each face rotation and each relevant slice rotation is a 4-cycle in each of the orbits. Finally, for odd $n\ge 5$, the combined permutation of off-center edge pieces, pieces on face diagonals, and pieces on face midlines at a fixed distance from the center is even. These parity restrictions are the only additional restriction; other than that, the orbits can be manipulated independently.
How does this change things? For the $n\ge 6$ case, all we have to do is implement the order $16\cdot 5$ piece as the product of a $16$-cycle, a $2$-cycle, and a $5$-cycle. Then all of the permutations we're working with are even, so there aren't any problems. In the smaller cases, we'll have to check.
For $n=2$, only the corners matter, and the highest order is $45$.
For $n=3$, we have corners and edge centers. With $17$ possible orders for the first and $31$ for the second, we can just scan all of the possible combinations. The highest LCM is $2520$, from order $45$ (3-cycle, 5-cycle, rotations in the 3-cycle don't add to zero) in the corners and order $56$ (4-cycle, 7-cycle, reflections in the 4-cycle don't add to zero) in the edges. This, however, doesn't pass the parity check; the combined permutation is odd, and there's no wiggle room to change that. That leaves us dropping down to the second-highest possibility of $1260$, possible in several ways ($45$ and $28$, $45$ and $84$, $35$ and $36$, $70$ and $36$). Three of those fail the parity check, leaving only the pair of order $45$ in the corners and order $28$ (Two 2-cycles, a 7-cycle, reflections in at least one 2-cycle don't add to zero) in the edges. Highest order $1260$, as previously noted.
For $n=4$, we have the corners, one orbit of off-center edges, and one orbit of off-center face pieces. With over a hundred possible orders in $S_{24}$, finding the least common multiple isn't the easiest search. Pruning everything with an order that divides another gets that down to a manageable total of $27$, and $5$ for the corners. Running it in the spreadsheet I've been using, the LCM is $765765=3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17$; $45$ in the corners, $11\cdot 13$ in the edges, and $7\cdot 17$ in the faces. Checking parity... all the pieces are odd permutations, so we'll actually have something of that order.
For $n=5$, we have corners, edge centers, three size-24 orbits, and two parity restrictions that cross orbits. The best I've got here is $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 23 = 281801520$; it could use an independent check, but I'm pretty sure it's right. The three size-24 orbits get us orders of $23$, $7\cdot 17$, and $11\cdot 13$ while the corners give us order $45$ and the edges give order $16$ (an 8-cycle and a 2-cycle, with an odd number of reflections in the 8-cycle). They're all odd permutations, so the parity checks pass.
For $n\ge 6$, as previously noted, we reach the maximum possible order $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23 = 5354228880$.
Best Answer
You can certainly achieve a rotation of side as a composition of moves of the other sides only. The most instructive way to see this is to take your favorite algorithm for solving the cube from an arbitrary position and figure out how to modify it such that it never turns the first side you solve. (Whether this is easy depends on what your favorite algorithm is, of course. My own usual one is almost there except for the initial-cross stage and the final turning of the top corners, both of which are easily replaced). Then turn your chosen side by one quarter, and use the modified algorithm to solve from that position. Whatever moves you make during the solve will add up to a "turn this side without turning it" combination.
Also, 5 is the minimum number of single-side turns that generate the group, because restricting movement to any set of 4 sides will either leave one edge unmovable or be unable to flip an edge to the opposite orientation.
A full set of relations between 5 basic quarter-turns would be absolute horrible, though.
If the generators are not required to be simple moves, we don't need as many as 5 of them. It's easy to get down to 3 quite complex operations:
Here there's some hope of being able to write down some reasonably natural relations, but I haven't tried to see whether the details work out.
I'm not sure if we can get down to two generators -- could the role of $\gamma$ be folded into $\beta$, perhaps? Edit: The next time the question came up (and I had forgotten this one existed) I tried anew and came up with a blueprint for a two-generator solution: Minimal generating set of Rubik's Cube group
One generator is of course not possible, since the group is not abelian.