If I have chased through the literature correctly, I think the answer to your question is "yes". Specifically:

Dyer--Formanek--Grossman showed that $\mathrm{Aut}^+(F_2)$ is isomorphic to $B_4/Z_4$, the quotient of the 4-strand braid group by its centre. (Bridson--Wade give a geometric proof here).

In their Theorem 1(i), Charney--Crisp state that $\mathrm{Out}(B_4/Z_4)\cong\mathbb{Z}/2\mathbb{Z}$. (Note that they denote $B_4$ as the Artin group $A(A_3)$.) They attribute the theorem to a 1981 paper of Dyer--Grossman.

In summary, $\mathrm{Out}(\mathrm{Aut}^+(F_2))$ has order 2, and hence $\mathrm{Aut}(\mathrm{Aut}^+(F_2))$ must be $\mathrm{Aut}(F_2)$.

$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Out{Out}$If I have chased through the literature correctly, I think the answer to your question is "yes". Specifically:

Dyer–Formanek–Grossman showed that $\Aut^+(F_2)$ is isomorphic to $B_4/Z_4$, the quotient of the 4-strand braid group by its centre. (Bridson–Wade give a geometric proof here).

In their Theorem 1(i), Charney–Crisp state that $\Out(B_4/Z_4)\cong\mathbb{Z}/2\mathbb{Z}$. (Note that they denote $B_4$ as the Artin group $A(A_3)$.) They attribute the theorem to a 1981 paper of Dyer–Grossman.

In summary, $\Out(\Aut^+(F_2))$ has order 2, and hence $\Aut(\Aut^+(F_2))$ must be $\Aut(F_2)$.