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I can think of at least one case where the answer is clearly no.

For example, if $F$ is a 3-dimensional compact hyperbolic manifold and the base space is any compact manifold, $E$ can't be hyperbolic.

The idea that that diffeomorphism group of $F$ is homotopy-discrete (this is a combination of work of Hatcher, Waldhausen and Mostow), having the homotopy-type of $Isom(M)$. So in this case the bundle $F \to E \to B$ has to have structure group a group of isometries, making $E$ a product geometry. Product geometries aren't hyperbolic, by the Margulis lemma (take elements of infinite order in $\pi_1 F$ and $\pi_1 B$ and you can construct a $\mathbb Z^2$-subgroup of $\pi_1 E$ with little effort).

I suspect an argument like this should work in more generality.

I can think of at least one case where the answer is clearly no.

For example, if $F$ is a 3-dimensional compact hyperbolic manifold and the base space is any compact manifold, $E$ can't be hyperbolic.

The idea that that diffeomorphism group of $F$ is homotopy-discrete (this is a combination of work of Hatcher, Waldhausen and Mostow), having the homotopy-type of $Isom(M)$. So in this case the bundle $F \to E \to B$ has to have structure group a group of isometries, making $E$ a product geometry. Product geometries aren't hyperbolic, by the Margulis lemma (take elements of infinite order in $\pi_1 F$ and $\pi_1 B$ and you can construct a $\mathbb Z^2$-subgroup of $\pi_1 E$ with little effort).

I suspect an argument like this should work in more generality.

edit: yes, this argument holds in greater generality. Since it's essentially a fundamental-group issue you can avoid the Hatcher + Waldhausen part of the argument above and appeal directly the Mostow. Even if the monodromy for your bundle isn't of finite order, it is up to homotopy by Mostow, provided the dimension of the fibre is $\geq 3$.

So I think that means the only case that has not been covered is when the fibre is 2-dimensional.

This isn't an answer. Humm...

I can think of at least one case where the answer is clearly no.

For example, if $F$ is a 3-dimensional compact hyperbolic manifold and the base space is any compact manifold, $E$ can't be hyperbolic.

The idea that that diffeomorphism group of $F$ is homotopy-discrete (this is a combination of work of Hatcher, Waldhausen and Mostow), having the homotopy-type of $Isom(M)$. So in this case the bundle $F \to E \to B$ has to have structure group a group of isometries, making $E$ a product geometry. Product geometries aren't hyperbolic, by the Margulis lemma (take elements of infinite order in $\pi_1 F$ and $\pi_1 B$ and you can construct a $\mathbb Z^2$-subgroup of $\pi_1 E$ with little effort).

I suspect an argument like this should work in more generality.

There are trivial answers yes if you don't demand compactness, for example the hyperbolic plane fibers by hyperbolic lines -- the orthogonal complement to geodesics. Similarly, higher dimensional hyperbolic spaces fiber, and there's unknown (google)'s 1-dimensional base example.

But if both the base and fiber are not simply-connected and everything is compact, then I think the answer is no. Take elements of infinite order in the fundamental group of the base and fiber respectively. Since the automorphism group of the fiber is finite, this gives you a $\mathbb Z^2$ subgroup of $\pi_1 E$ which contradicts Margulis.

This isn't an answer. Humm...