You can find such a map which is degree one. The cut number $n$ of a closed 3-manifold $M$ is the maximal number of 2-sided disjointly embedded surfaces $\Sigma_1,\ldots,\Sigma_n$ which do not separate $M$. For each connected surface $\Sigma_i$, we can make a degree one map to the 2-sphere by taking a spine of the surface (a embedded graph whose complement is a disk) and crushing it to a point. Do this for each component of the surfaces $\Sigma_i$ in the 3-manifold mapping to $S^2\times \ast$ for each factor of $\#_n S^2\times S^1$. The complement of these 2-spheres is a $2n$-punctured 3-sphere. There is no obstruction to extending the maps of $\cup_n \Sigma_i$ to $\cup_n S^2\times \ast$ to a map $M-\Sigma_i \to \#_n S^2\times S^1 - \cup_n S^2 \times \ast$, and get a degree one map from $M$ to $\#_n S^2\times S^1$. Now, we note that $\Sigma\times S^1$ has cut number at least $g$, by taking $g$ disjoint non-separating curves on $\Sigma$ and crossing them with $S^1$.