Lemma IX.4.2 of Subsystem's of second order arithmetic Simpson claims that $\textbf{WKL}_0$ is sufficient to prove that every consistent theory has a saturated model. I will summarize the proof:
Let $S$ be a consistent $L$ theory and add to the language of $S$ a sequence of new constants $C=(c_n)_{n\in\mathbb{N}}$. A sequence $(\varphi_i)_{i\in\mathbb{N}}$ of $L\cup C$ formulas is acceptable if it only uses one free variable and finitely many constants. We fix an enumeration of the recursive acceptable sequences $((\varphi_{n,i}:i\in\mathbb{N}))_{n\in\mathbb{N}}$. Let $S'$ be $S$ plus the Henkin axioms:
\begin{equation*}
\exists x \bigwedge_{i\leq j}\varphi_{n,i}(x)\rightarrow \bigwedge_{i\leq j}\varphi_{n,i}(c_m)
\end{equation*}
For a suitable $m$.
Using Lindembaum's Lemma we can find a complete extension of $S'$ and then the term model is the wanted recursively saturated model.
My issue with this proof is that it seems that $\textbf{WKL}_0$ is too weak to ensure that there is an enumeration of the recursive acceptable sequences of formulas but rather that $\textbf{ACA}_0$ is sufficient.
The only fix that comes to mind is that we fix an enumeration of the partial acceptable sequences of formulas and the theory $S'$ will in this case be recursively enumerable and then one uses Craig's trick to get a recursive theory.