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 $S^1$.
Remark on extending maps: Take a vertex in $M$, and connect it by $2n$ arcs to the surfaces $\Sigma_i$, one for each side of $\Sigma_i$ (which is possible by the non-separating assumption). Then the boundary of a tubular neighborhood of this graph and $\cup_i \Sigma_i$ gives a connected surface. We map do a similar thing in $\#_n S^2\times S^1$ with the surfaces $\cup_n S^2\times \ast$, where the boundary of the tubular neighborhood is a 2-sphere which bounds a ball. Extend the map on surfaces to a map on graphs, and then of a tubular neighborhood (getting a degree one map), then we may extend the map over the ball (just crush the degree one map of the surface to $S^2$ to a point to extend over the ball).
