In certain very special cases, I think this can be answered. In particular, if $f:\Sigma_g \twoheadrightarrow S_n$ is onto, $n\geq 4$ and $g \gg 0$, then Theorem 6.20 of Dunfield-Thurston implies that the map $f$ is determined up to the action of the mapping class group by the image $f_\ast:H_2(\Sigma_g) \to H_2(S_n) \cong \mathbb{Z}/2$ (when $n\geq 4$). In the case that the image is zero, then $f$ can be chosen to factor through a handlebody of genus $g$, and since the map factors through a free group, this will lift to $B_n$ since there is no obstruction.
If $f_\ast$ is non-trivial, then there will be a lift iff $\mathbb{Z}\oplus \mathbb{Z}/2 = H_2(B_n)\twoheadrightarrow H_2(S_n)=\mathbb{Z}/2$ is surjective. I'm pretty sure that this is true, and it should be represented by a torus whose fundamental group is generated by $\sigma_1, \sigma_3$ in the standard braid group generators. One need only check that this torus maps homologically non-trivially into $H_2(S_n)$, which I think follows from the presentation of the double cover of $S_n$.
If $f$ is not onto, then one could still attempt to apply Theorem 6.20 to its image. Let $f(\Sigma_g)= H < S_n$. Then Theorem 6.20 implies that for $g$ large enough, $f$ is classified up to the mapping class group by the image of $f_*: H_2(\Sigma_g)\to H_2(H)$, up to the action of $Out(H)$. Let $\tilde{H} = p^{-1}(H)$ be the preimage of $H$ in $B_n$. If $p_{|\ast}: H_2(\tilde{H}) \to H_2(H)$ is not onto (again, up to the action of $Out(H)$), then one could find a counterexample. I think there's a good chance of such a subgroup existing.