May be a possible way to solve the problem is to set up it in the context of closed relations instead of functions, by putting $R = g^{-1} \circ f : B^n \to B^n$, and assuming for instance $f(B^n) \subset g(B^n)$. So, we are looking for $x \in B^n$ such that $x \in R(x)$. 


In shape theory there are defined the so-called multihomotopy groups, which generalize the classical homotopy groups by considering multi-valued loops instead of the single-valued ones. See http://plms.oxfordjournals.org/content/s3-69/2/330 and http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.jmsj/1226598673

Under nice circumstances the multihomotopy groups of spheres are $\tilde\pi_n(S^n) \cong \Bbb Z$, while for the $n$-disk are trivial. 


One can try to adapt the proof of Brouwer's fixed point theorem in this setting.


So, assuming by contradicition that $f(x) \ne g(x)$ for all $x \in B^n$, one can consider the relation $T : B^n \to S^{n-1}$ where $T(x)$ is the intersection between $S^{n-1}$ and the half-lines starting from points of $R(x)$, and passing through $x$. It follows that $T$ is the identity on $S^{n-1}$, so $T$ is a multi-valued retraction $B^n \to S^{n-1}$, and this contradicts the fact that $\tilde\pi_{n-1}(S^{n-1})$ is not trivial. 

I think that there are many details to fill, and that one should give the "nice" conditions under this approach works (for instance, assuming that $g^{-1}(y)$ is countable for all $y \in B^n$ does suffice?) but this could be an interesting application of multihomotopy groups.