To expand on Marty's brief comment, self-duality is not possible for any $p$ and for many if not most finite groups. (Also, your tag 'lie-groups' is not meaningful here, since a finite group of rational points requires starting with an algebraic group. It does turn out after a lot of work that semisimple complex Lie groups are essentially the same as semisimple algebraic groups over $\mathbb{C}$, but for this question you need tags 'finite-groups' and 'algebraic-groups' rather than 'finite-fields' and 'lie-groups'.)
Starting with a classical algebraic group $H$ on your list (or even a Spin group), the question makes sense over any algebraically closed field such as an algebraic closure of a finite field. In this situation, the adjoint representation of $H$ has been thoroughly explored in older work by Hogeweij and Hiss. As in characteristic 0, $H$ has a finite center equal to the kernel of the adjoint representation. But the Lie algebra sometimes has a center of positive dimension, e.g., $\mathfrak{sl}_n$ when the characteristic $p$ divides $n$. Aside from this possibility, the modular representations of an arbitrary finite group are typically not self-dual---indeed, the module obtained is often indecomposable but not irreducible. There are lots of examples for finite classical groups $G$ of Lie type, even when they are embedded in the rational points of a suitable $H$ over a finite field.
An easy example to keep in mind involves the principal series representations of the finite group of Lie type $G=\mathrm{SL}_2(\mathbb{F}_p)$: here you typically have two composition factors, with a nonsplit extension, and the dual module has the same two (self-dual!) composition factors but arranged in the opposite order.
Note that Jantzen has worked out many details for such principal series in a more general setting here.