Let $S$ be a complete hyperbolic surface of finite area, $f: S\to S$ be a pseudo-Anosov homeomorphism which lies in a torsion-free finite index subgroup $\Gamma$ of the mapping class group $Mod_S$ (for instance, $f$ acts trivially on mod-3 homology of $S$). Let $M_f$ denote the mapping torus of $f$: Such manifolds are known to be hyperbolic and every complete finite volume hyperbolic 3-manifold is finitely covered by such a mapping torus. Hence, in a sense, the above examples are "generic."

Proposition. There exists a smooth quasi-projective surface $X$ and a proper $\pi_1$-injective embedding $M_f\to X$.  

Here is a sketch of the proof. Let ${\mathcal M}={\mathcal M}_{\Gamma}$ denote the quotient of the Teichmuller space of $S$ by $\Gamma$. Then ${\mathcal M}$ is a smooth quasi-projective variety. Hence, applying a form of Lefschetz hyperplane section theorem for quasiprojective varieties, we find a properly embedded smooth quasi-projective curve $C\subset {\mathcal M}$ such that $\pi_1(C)\to \pi_1({\mathcal M})$ is surjective. In particular, there exists a loop (possibly non-simple) $c$ in $C$ which represents the conjugacy class of $f$ in the mapping class group. Let $p: Y\to {\mathcal M}$ denote the universal curve, whose fibers are diffeomorphic to $S$. Then $Z:=p^{-1}(C)$ is a smooth quasi-projective surface. The pull-back $W\to C$ of $p$ to $c$ is diffeomorphic to our mapping torus $M_f$. By the construction, the natural map $M_f\to W\to Z$ is $\pi_1$-injective. However, it need not be injective because the loop $c$ need not be simple. Luckily, there is a finite-sheeted covering $C'\to C$ and a simple loop $c'\subset C'$ such that the image of $c'$ in $C$ is freely homotopic to $c$. (This is a consequence of the so called LERF property of surface groups.) Hence, after replacing $C$ with $C'$ and $p: W\to C$ with $p': W'\to C'$ which is the pull-back of $p$ to $C'$, we obtain the required 2-dimensional smooth quasi-projective variety $X=W'$ and the $\pi_1$-injective proper embedding $\iota: M_f\to X$.  

I do not see a reason to expect $\iota$ to be an isometric embedding for any complete Kahler metric on $X$.