If you consider the unstickered cubies and their orientations to be distinguishable, I believe there are \( 54!/(9!^6) \) possible stickerings. If you consider the orientation of the cube not to matter, it would be approximately 24 times fewer (although I believe the exact number to be rather tricky to calculate). As with the 4x4x4 cube with its indistinguishable pieces, these arrangements of indistinguishable stickers do not form a group. Also S56/S9 (or S54/S9, if that's what you meant) is not a group, since S9 is not a normal subgroup of S56 (or S54). Even though those coset spaces are not groups, we may still be able to talk about an isomorphism, provided the the number of elements matched (although that is not the case here either).Hm, so I have a conjecture which I'm not keen to focus on because it's not very interesting, but just out of curiosity is it true that the group of all the states reachable by peeling off the stickers and sticking back on them anywhere is isomorphic to S56/S9? I think the cardinality of that group is right and there's what appears to me to be an intuitive isomorphism but I haven't tried proving it at all.
FWIW by S56 I mean the Symmetric Group on 56 letters.
It might also be a somewhat interesting question to determine the total number of possible stickerings if we allow stickerings to be considered equivalent if one can be converted to the other by some element of <U,x,y>. One might also allow stickerings to be considered equivalent if one can be converted to the other by disassembling and reassembling the cube. (I am assuming the stickers to have square symmetry, are a solid color with no markings, and are divided into 6 sets of 9 indistinguishable stickers.)