The answer is negative even for smooth projective varieties over $\mathbb C$: a counterexample is constructed in http://arxiv.org/abs/1609.06391.  The example is a smooth, uniruled sixfold of Picard rank somewhere in the high 30s.

Here's the basic idea.  One should try to find a variety $X$ whose automorphism group is something like a free group. It's possible to rig things so that there's a point $x$ on $X$ whose stabilizer in $\operatorname{Aut}(X)$ is a non-finitely generated subgroup; free groups have lots of these.  If you blow up $x$, the automorphisms of $X$ that lift to the blow-up are precisely the ones that fix $x$.  One also needs to check that the blow-up doesn't have any automorphisms other than the ones lifted from $X$, but this isn't too hard.  Things don't work out quite this cleanly in the linked example, but this is the strategy.

There's also the issue of how to actually prove the automorphism group isn't finitely generated.  We've arranged that the "obvious" automorphisms we can write down are a non-finitely generated group, but we might worry that there are other automorphisms we don't know about.  Maybe we actually just found a non-finitely generated subgroup of some larger finitely generated group.  The solution is to arrange that there's a rational curve $C$ on $X$ and prove that every automorphism of $X$ must fix $C$, and in fact restrict to $C$ as a map of the form $z \mapsto z+c$.

This means we have a restriction map $\rho : \operatorname{Aut}(X) \to \operatorname{Aut}(C)$, and the image lies in the abelian subgroup of translations fixing $\infty$.  We construct explicit automorphisms $\mu_n$ (for every $n$) whose restrictions to $C$ are the maps $z \mapsto z + 1/3^n$.  The means the image of $\rho$ is an abelian group containing $\mathbb Z[1/3]$, so it's not finitely generated.  This means that $\operatorname{Aut}(X)$ isn't finitely generated either, though I don't know exactly what it is.  If I had to guess, it's probably free on countably many generators.