I'm interested in a solution to this problem. Anyone have one?

If we cut the corners (8 black triangles) of a cube with 6 white faces, we obtain a polyhedron satisfying the coloring condition but the corners are a tad far from the center. l think this is representative of the dilemma between circumscribing around a sphere and having on average n>m for white n-gons and m-gons to satisfy the coloring condition. (This is because each edge is shared by a white and a black, but there are more black faces). Consider a characteristic c(n) of a n-gon = E/D, where E is the sum of all edges, and D is the sum of distances from vertices to an interior point P which is tangent to the sphere. Perhaps we could derive a cobtradiction between the range of c(n) and c(m), while the entire sets of black e's and d's are identical with those of white. I am driving and cannot really work this out now when a kid called me about this, but I will think about it later....

The result is probably false. Consider the Herschel graph, https://en.wikipedia.org/wiki/Herschel_graph. It can be realized as the polyhedral graph of some polyhedron $Q$. We can assume $Q$ is inscribed in a sphere $S$: just take a large sphere and project $Q$ onto it. Now invert $Q$ with respect to the sphere $S$ to get a polyhedron $P$ circumscribed about $S$. From this construction, the adjacency relations among the faces of $P$ are captured by the Herschel graph. Because this graph has 11 vertices and it is bipartite, one can color 6 faces black such that no two black faces are adjacent.

This seems to amount to asserting that every convex polyhedron had an inscribed and circumscribed topological equivalence. An immediate logical consequence is that the cube octahedron is also circumscribed -- just project its conjugate to the sphere and invert back. In fact I think the spherical inversion is only guaranteed to work for Platonic (plus a select set of) polyhedra; correct me if I am wrong.