I believe that projective pearls must be free.
As noted in the comments, a pearl epimorphism $f\colon (G,S)\to (H,T)$ is in fact surjective onto $H$ (though not necessarily onto $T$).
Let $(G,S)$ be a projective pearl. Let $F_G$ be the free group on the underlying set of $G$ (say, with free generators $\{x_g\mid g\in G\}$, and let $\pi\colon F_G\to G$ be the projection obtained by mapping the free generator $x_g$ of $F_G$ to the element $g\in G$. Letting $T=\{x_s\in F\mid s\in S\}$, we have that $(F_G,T)$ is a pearl, and $\pi\colon (F_G,T)\to(G,S)$ is a pearl homomorphism which is onto $G$, and hence an epimorphism. Since $(G,S)$ is projective, the surjection splits, so we have a pearl homomorphism $k\colon (G,S)\to (F_G,T)$ such that $\pi k=\mathrm{id}$. In particular, $\pi k\colon G\to G$ is the identity morphism, so $k$ is one-to-one. So $G$ is isomorphic to a subgroup of the free group $F_G$, and hence $G$ is itself free.