Can noncomputable sets be distinguishable in $RCA_0$?

Yes.

Claim. There is a noncomputable $\Delta^0_2$ set $X$ which is distinguishable from every set it computes.

To ensure distinguishability, we must create a machine $\Phi$ such that for every functional $\Psi_e$ with $\Psi_e^X$ total and $\Psi_e^X \neq X$, there is a pair $(\sigma, \tau)$ with $\sigma \prec X$, $\tau = \Psi_e^\sigma$, $\tau \neq \sigma$, $\Phi^{\sigma\oplus\tau}(0) = 0$ and $\Phi^{\tau\oplus\sigma}(0) = 1$. So we have a requirement for every $e$, and we wait until we see a pair $(\sigma, \tau)$ that we like and then enumerate such axioms into $\Phi$. But when we enumerate the axioms, we are forever denying the possibility that $\tau$ will be an initial segment of $X$. And we will need to occasionally change $X$ to ensure noncomputability.

So we simply arrange that the $\tau$ are chosen long, so that $\sum_{(\sigma, \tau)} 2^{-|\tau|}$ is small. Then there will always be strings available to change to. Now arrange the above distinguishability requirements into a finite injury construction along with standard noncomputability requirements.

I suspect that with some care, this could be modified to make $X$ c.e..

Now consider the model of $RCA_0$ $(\omega, \{Y : Y \le_T X\})$. This has a distinguishable, noncomputable set.

Extending D. Turetsky's answer: If $\mathcal{M}$ is a Turing ideal containing no PA degree, then every $\Delta_2$ degree in $\mathcal{M}$ contains a set that is computably distinguishable within $\mathcal{M}$.

Lemma. If $\mathcal{M}$ contains no PA degree, then $\mathcal{M}$'s version of the Cantor space $2^\omega$ is computably homeomorphic to $\mathcal{M}$'s version of the Baire space $\omega^\omega$.

Proof. Fix a computable binary tree with no path, and name its leaf nodes $\tau_1,\tau_2,\ldots$. Every $f \in 2^\omega$ is a concatenation of infinitely many $\tau_{n_1}\tau_{n_2}\cdots$ and through this maps to the string $n_1n_2\cdots \in \omega^\omega$.

Proof of main claim. Fix a $\Delta_2$ set $D$ in $\mathcal{M}$. Construct a computable tree $T \subseteq \omega^{<\omega}$ whose unique path $P$ is $D$'s computation function, i.e., the map taking $n$ to the least $s$ for which $D_s\cap[0,n] = D \cap [0,n]$. Transform $T$ using the Lemma into a computable $T^* \subseteq 2^{<\omega}$. Then $T^*$ has a unique path $P^* \equiv_T P \equiv_T D$, and this $P^*$ is distinguishable by the procedure which takes $X,Y$ and checks to see if their initial segments are all in $T^*$. $\square$

I'd be curious to know whether your proof of distinguishable-equals-computable really needs $WKL$ in the form of a $PA(X)$ degree, or whether it can be finessed down to a $PA$. It would narrow the gap if you're looking for a reversal.