In general, even if the fibers of $\phi$ are connected and the map is proper, there need not be an almost complex structure on $B^{2k}$ such that the differential of $\phi$ is complex linear. (I assume that you meant to assume that the kernel of the differential of $\phi$ is preserved by $J$, not that $J$ acts trivially on it, which doesn't make sense unless the kernel is $0$.)
Meanwhile, if there is an almost complex structure on $B^{2k}$ for which the differential of $\phi$ is complex linear, then, yes, that almost complex structure has to be integrable.