Yes. Any finitely generated module in category O is a quotient of a finite generated projective, and thus a quotient of a module with a finite Verma module filtration. Thus, any infinite dimensional module tensored with a Verma module has weight multiplicities that grow too fast (faster than a constant times the Kostant partition function), so it can't be in category O.

In response to Jim's challenge, I think I've found the easiest approach a bit belatedly: any module in category $\mathcal{O}$ is finitely generated over $U(\mathfrak{n}_-)$, and any Verma module is free of rank 1 over it. Thus, if $M$ is infinite dimensional, and $N$ a Verma module, the tensor product $M\otimes N$ is free of infinite rank over $U(\mathfrak{n}_-)$, and thus not in $\mathcal{O}$.