$\DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\SL}{SL}
\DeclareMathOperator{\Aut}{Aut}$
For clarity, let's assume that $V$ is a module on which $G$ acts by a faithful representation $D\colon G \to \GL(V)$. Then "every element of $H$ is induced by $\SL(V)$" means that for every $h\in H$ there is some $S\in \SL(V)$ such that $D(g^h) = D(g)^S$ for all $g\in G$.

The exact answer will depend on the assumptions on the module $V$ and the ground field:

When $V$ is an irreducibe, faithful module over the complex numbers
(or any algebraically closed field of characteristic $\neq p$), then
$H=\Aut_{\SL(V)}(G)$.

Proof. An irreducible, faithful character of $G$ vanishes outside the center, and is determined by its values on the center. Thus the representation $g\mapsto D(g^h)$ is similar to $D$ for $h\in H$, and so there is $S\in \GL(V)$ with $D(g^h) = D(g)^S$ for all $g$. As the field is algebraically closed, we can multiply with a suitable scalar to get $S\in \SL(V)$. This shows $H\subseteq \Aut_{\SL(V)}(G)$.

Conversely, when $V$ is absolutely irreducible, then every element in the normalizer of $D(G)$ in $\SL(V)$ (or $\GL(V)$) induces an element of $H$, since the center of $G$ is mapped to the center of $\GL(V)$ by $D$. Thus $\Aut_{\SL(V)}(G) \subseteq H$.

In all other cases, both inclusions may fail:

We still have $H\subseteq \Aut_{\SL(V)}(G)$, when $V$ is not irreducible, but all irreducible constituents are faithful. On the other hand, if we allow for non-faithful (=linear) constituents, then $h\in H$ may map these to other constituents, and so $D$ and $D^h$ can not be similar.

Also, when the field is not algebraically closed, but $V$ irreducible (even absolutely irreducible), then $D$ and $D^h$ are similar, but maybe not in $\SL(V)$. This happens in the exponent $p^2$ case, and also in the case where $G$ has order $3^3$ and exponent $3$.

If $V$ is not absolutely irreducible, then an element in $\Aut_{\SL(V)}(G)$ may induce a non-trivial action on the center of $G$. (Whether this can happen is determined by the restriction of $V$ to the center.)

And finally, in characteristic $p$ it's not true, as said by Derek Holt. (In characteristic $p$, the only simple module is the trivial module, so we would have to allow arbitrary modules here.)