The Discrete Hairy-Ball Theorem

I am not sure I fully understand the question, but you should know the following classical discrete version of the hairy ball theorem:

Let $P \subset \Bbb R^3$ be a convex polytope. We say that an orientation of edges of $P$ is balanced if every vertex has at least one ingoing and one outgoing edge (i.e., the oriented graph has no sinks and no sources). Then the edges of at least two faces of $P$ form oriented cycles.

The proof is completely straightforward and goes exactly the way you think it should go (follow the arrows!). This result has been rediscovered a number of times. The earliest place where I found this result is here: L. Glass, A combinatorial analog of the Poincaré index theorem, J. Combin. Theory Ser. B 15 (1973), 264–268. See also Ex. 25.15 in my book for the context, and connections with other related results (sorry for the self-promotion).

By interpolating "legal configuration" if it were to exist, I think you can obtain a nowhere vanishing, continuous vector field on $S^2$. Suppose that for some $n$ there is a legal configuration on the $n\times n \times n$ cube. For each 1x1 face, put a vertex at the center. Then connect vertexes whose 1x1 face's touch. (basically something like a dual graph, but I don't know what the real terminology is).

Doing this, you get $(n-1)^2$ squares on each face of the Rubik's cube, $n-1$ squares on each edge of the Rubik's cube, and one triangle for each vertex of the cube. Now, put the arrow from the 1x1 face at the associated vertex. For each square or triangle we can now linearly interpolate to get a vector field over the whole thing. These will patch back together to form a continuous vector field on $S^2$ because on the lines they are glued along, the value on each piece is linear interpolation between the same two vectors. Thus, basically it remains to check that given a "legal configuration" you cannot interpolate to a zero vector.

The squares are not too bad, because the most that two of the vectors being interpolated can be off by is $90^\circ$, so you can't get a zero vector.

The triangles are a little trickier because things get twisted around, but if you try to write down the possible cases, you can see that there is basically only one type of "legal" corner configuration, and it doesn't interpolate to a zero vector.

This seems to show that you don't even need to assume anything special about corner squares, you can allow them to point in the 8th illegal position, and there are still no "legal configurations."