I'll try. Take a genus 2 Riemann surface $S$, and embed it (differentiably) in $\mathbb{R}^3$ so that it's symmetric with respect to a 2-plane $\Pi$. Consider the intersection $S\cap \Pi$: choosing the embedding suitably, such intersection is the union of three circles $\alpha$, $\beta$ and $\gamma$. The complement of $\alpha \cup \beta \cup \gamma$ is a union of two noncompact surfaces $U_0$ and $V_0$, each homeomorphic to a sphere whith 3 holes, and each hole (within both $U_0$ and $V_0$) is bounded by the circles $\alpha$, $\beta$ and $\gamma$. If you thicken $U_0$ and $V_0$ you'll obtain an open cover {$U,V$} of $S$, such that $U\cap V$ is a union of thickenings of $\alpha$, $\beta$ and $\gamma$ which we call $W_{\alpha}$, $W_{\beta}$ and $W_{\gamma}$.
Now, a bundle on $S$ with fiber $F$ is encoded (up to isomorphism) by (the homotopy class of) a map
$\varphi:W_{\alpha}\cup W_{\beta} \cup W_{\gamma} \to \mathrm{Diff}(F)$.
Take as $F$ your favourite hyperbolic surface, and suitably represent it in $\mathbb{R}^3$ as above, symmetrically with respect to a 2-plane $\Pi$.
Define $\varphi$ to be the identity on the connected components $W_{\alpha}$ and $W_{\beta}$ and the symmetry with respect to $\Pi$ on the connected component $W_{\gamma}$.
Let $X$ be the total space of the resulting (differentiable) bundle. If I'm not mistaken, $X$ cannot have a complex structure because it's not orientable.