Hmmm, it seems the answer is no.
Fix any nonhyperarithmetic real $x$ so that the $\Sigma^1_1(x)$ set $$A_x=\{y\mid \forall n \in \mathscr{O}^x\forall m\in \mathscr{O}^y(\mathscr{O}^x_n\cong \mathscr{O}^y_m\rightarrow x\not\leq_T y^{(|m|)})\}$$ is not empty, where $\mathscr{O}^x$ is Kleene's $\mathscr{O}$ relative to $x$ and $\mathscr{O}^x_n$ is $\mathscr{O}^x$ restricted to $n$ and $|m|$ is the $y$-recursive ordinal coded by $m$, and $y^{|m|}$ is the $|m|$-th Turing jump relative to $y$.
To see that $A_x$ is $\Sigma^1_1(x)$, just notice that the isomorphism between $\mathscr{O}^x_n$ and $\mathscr{O}^y_m$ must be hyperarithmetic in $x\oplus y$.
To see the existence of such $x$, just let $x$ be a real Turing computing all hyperarithmetic reals but $\omega_1^x=\omega_1^{CK}$. Then any $\Delta^1_1$-random real $y$ with $\omega_1^y>\omega_1^{CK}$ must belong to $A_x$.