It is a theorem of Gromov and Lawson, also Schoen and Yau, that no closed orientable three-manifold which contains an aspherical factor in its prime decomposition can admit a metric of positive scalar curvature, see Theorem IV.6.18 of Spin Geometry by Lawson and Michelsohn. In particular, as the three-manifold you're interested in is aspherical, it does not admit a metric of positive scalar curvature (irrespective of the nature of the fibers).
In fact, thanks to the solution of the elliptisation conjecture, we now know that a closed orientable three-manifold admits a metric of positive scalar curvature if and only if its prime decomposition contains no aspherical factors.
A natural question to ask is whether your manifold can admit a metric of non-negative scalar curvature. In the absence of positive scalar curvature metrics, such a metric is Ricci-flat, and on a three-manifold, a Ricci-flat metric is flat. If $M$ admits a flat metric, it is finitely covered by $T^3$ and hence $b_1(M) \leq 3$. As $M$ is an orientable circle-bundle over $\Sigma_{\gamma}$, the Gysin sequence tells us that $\pi^* : H^1(\Sigma_{\gamma}) \to H^1(M)$ is injective, and hence $\gamma = 1$. If the Euler class of the bundle is non-zero, then $M$ is finitely covered by the Heisenberg manifold $H(3, \mathbb{R})/H(3, \mathbb{Z})$ which does not admit a flat metric. On the other hand, if the Euler class is zero, then $M = T^3$ which certainly does.