Maps of surfaces to CAT(0) cube complexes Let $C$ be a locally CAT(0) cube complex.  Let $f\colon \Sigma_g \rightarrow C$ be a continuous map from a closed oriented genus $g \geq 1$ surface to $C$.  Is it always possible to find a cube complex structure on $\Sigma_g$ such that $f$ is homotopic to a map taking cubes to cubes?
 A: This is not an answer but instead is a "story" to show that the answer is perhaps "no".
Suppose that $C$ is the given locally CAT(0) cube complex. Suppose that $S$ is the given surface of genus $g \geq 1$. Suppose that $f \colon S \to C$ is the given continuous map.
Suppose, as a very special case, that $f_* \colon \pi_1(S) \to \pi_1(C)$ is injective.   Then we homotope $f$ to be transverse to the cell structure of $C$, use Stallings-like techniques to homotope $f$ into good position, and pull-back a good structure on $S$.  (There are of course many details to fill in here.  The most troubling is that Stallings wants $f$ to be transverse, while the original question wants the image of $f$ to lie in the two-skeleton.)
Suppose more generally that $f_*$ is not injective.  If there is a simple closed curve in the kernel, then we can crush that, send it someplace standard, and then induct.
That is, this sort of inductive plan of attack seems (?) to need a version of the simple loop conjecture, where the target is a locally CAT(0) cube complex.  Unfortunately the simple loop conjecture is surely (?) false in high dimensions, even when the target is a very nice space (say a hyperbolic four-manifold).

As HJRW points out, we do not need high dimensionality to reach a possibly concerning place. A lovely example of the simple loop conjecture, when $S$ is a genus $g$ surface and $C$ is the product of a pair of rank $g$ graphs, appears on pages 85 and 86 of Stallings paper How not to prove the Poincaré conjecture.  This connection deserves to be better known!
