I don't really know much about perverse sheaves, but the answer for the first general question is "No", I think, even for curves.

Here is a simple example for $\bar{\mathbf{Q}}_{\ell}$-sheaves, which is the context I understand a bit. Take a Kummer sheaf on the multiplicative group over a finite field associated to a non-trivial character $\chi$ (often denoted $\mathcal{L}_{\chi}$), and define a perverse sheaf $F$ on the projective line as the middle-extension of this sheaf to the projective line (suitably normalized by shifting); this is just the extension by zero. Let $G$ be its dual; it is the perverse sheaf associated in the same way to $\mathcal{L}_{\bar{\chi}}$. Then $F\otimes G[-1]$, unless I misunderstand the definitions, is the extension by zero of the lisse sheaf $\mathcal{L}_{1}$ on the multiplicative group, i.e., the extension by zero to the projective line of the trivial rank $1$ sheaf. This is not perverse (because it would be simple perverse, being of rank $1$, and therefore, because of the classification of these and because it is not punctual, it would be the middle extension of its restriction to the multiplicative group, but the latter is the trivial sheaf on the projective line).