Edit: The answer is wrong - the comment gives a counterexample.
It seems Jason's comment gives the only exceptions: assume $d$ is prime to $p$ and $\deg(f) \geq 2$. Consider the factorization of $f$ in the algebraic closure of your field $\mathbb{F}$, say $f = \prod_{i=1}^e (x-a_i)$. Since $f$ is irreducible, none of these $a_i$ are in $\mathbb{F}$ (this is where we use $\deg(f) \geq 2$) and since it is separable, $a_i \neq a_j$ for $i \neq j$. Then
$f(x^d) = \prod_{i=1}^e (x^d-a_i) = \prod_{i=1}^e \prod_{j=1}^d (x-a_{ij})$
where $a_{ij}$'s are $d$-th roots of $a_i$. None of these $a_{ij}$'s are in $\mathbb{F}$ and none of them are equal. So $f(x^d)$ is irreducible as well.