$A\otimes_{\mathbb C}B$ is finitely generated as a $\mathbb C$-algebra. Does this imply that $A$ and $B$ are finitely generated?

Yes, it does, as long as $A$ and $B$ are both not the zero ring (obviously $A\otimes 0=0$ is finitely generated for any $A$). Choose a finite set of generators of $A\otimes_\mathbb{C} B$; each of these is a finite sum of tensors $a\otimes b$. Let $A_0\subseteq A$ be the subalgebra generated by all the $a$'s appearing in these tensors. Then $A_0$ is finitely generated, and we see that the natural map $A_0\otimes_\mathbb{C} B\to A\otimes_\mathbb{C} B$ is surjective (since its image contains all of the tensors $a\otimes b$ in our generators). This means $A/A_0\otimes_\mathbb{C} B=0$, so as long as $B\neq 0$, we must have $A/A_0=0$ and so $A_0=A$. Thus $A$ is finitely generated. By the same argument, $B$ is also finitely generated.

This argument clearly works with $\mathbb{C}$ replaced by any field. Much more generally, a similar argument shows that if $R$ is any base ring and $A$ and $B$ are $R$-algebras such that $B$ is faithfully flat over $R$, then if $A\otimes_R B$ is finitely generated as a $B$-algebra (in particular, if it is finitely generated as an $R$-algebra), then $A$ is finitely generated as an $R$-algebra.