Hence the original problem is equivalent to the existence of a matching in this bipartite graph. A necessary and sufficient condition for such a matching to exist is given by Hall's Marriage Theorem. We need to check that for any set of $k$ cubes, the cardinality of the union of 'visible orientations' linked to these cubes is greater than or equal to $k$. But $k$ cubes give rise to $24k$ edges, which give rise to at least $24k/(n-5)$ 'visible orientations', which is greater than or equal to $k$ for $n \leq 29$.
So there is a strategy for $n = 29$, and conversely this is the best possible just by comparing the number of vertices on either side.
Update: An explicit strategy was requested, so here is an adaptation of the standard card trick strategy:
Let's assume we're working with numbers from $0$ to $28$ for convenience.
The cube-orienter adds up the numbers on all the faces of the cube modulo $6$. Call the result $i$, with $0 \leq i \leq 5$. Now if $i=0$, put the smallest face face-down, if $i=1$, put the second smallest face-down, and so on. Later on, the guesser will be able to add up all the visible faces modulo $6$, and a little thought shows that the hidden face will be congruent to the negative of this modulo $6$, provided the guesser renumbers the unseen numbers from $0$ to $23$. This means the guesser will know the answer is one of $4$ cards, and these remaining $4$ degrees of freedom can be communicated by the $4$ possible rotations the cube-orienter can leave the cube in (with a fixed face down). E.g. of the side-faces, the largest can be pointing left/forward/right/backward.