I am assuming the cubical polytopes are convex - otherwise it's easy to come up with counterexamples - say cubulated tori.
In this case the distinction can be framed as follows. Say you have a cubical convex polytope - this is homeomorphic to a ball so its boundary always gives you a cubical sphere. So the set of cubical polytopes are contained in the set of cubical spheres (in the sense of combinatorial equivalence of their face lattice). The more interesting question is whether each cubical sphere is combinatorially equivalent to the boundary of a convex cubical polytope.
In three dimensions (or equivalently for $2$-spheres) cubical polytopes and cubical spheres coincide. Given any cubical $2$-sphere, its $1$-skeleton is a 3-connected planar graph so we can use Steinitz's Theorem to construct a polytope with the same combinatorial type.
Things get trickier in higher dimensions. I do not know of an explicit example of a non-polytopal $3$-sphere in the cubical case but the simplicial case has been studied extensively. There are many examples of simplicial $3$-spheres that are not combinatorially equivalent to the boundary of any $4$ simplicial polytope. The typical construction uses a knot on $3$ edges inside the simplicial sphere. If the knot is non-trivial the corresponding sphere will be non-shellable and thus not polytopal since every polytope is shellable. See Theorem 1 and 2 here. In the same paper Lutz gives a small example of a non-polytopal sphere on $13$ vertices. I think one could use the same ideas starting from Furch's knotted cube to get a non-shellable cubical 3-sphere.