Włodzimierz Holsztyński suggests, in comments to my other answer, a modification/subcategory of pearls: call a morphism of pearls $f\colon (G,S)\to (H,T)$ sharp if and only if $f(S)=T$. Now consider the sharp subcategory to be the category of pearls with sharp morphisms between them (and similarly for abelian pearls). Now asking if in this subcategory free pearls might be projective, and if there are projectives that are not free. I am writing a separate answer because the previous one was already a bit long, and this deals with different concept of morphisms.
As far as epimorphisms in the sharp categories go, my arguments given for the general category do not work. I do not yet know if all epimorphisms in the sharp category have underlying group homomorphism that is surjective. However, this still holds for the abelian pearls. In addition, it is still the case that neither sharp subcategory has projectives.
Theorem. Let $f\colon (G,S)\to(H,T)$ be a sharp morphism of abelian pearls. Then $f$ is an epimorphism in the sharp subcategory of abelian pearls if and only if $f\colon G\to H$ is a surjective morphism of abelian groups.
Proof. Assume $f$ is surjective. Let $(K,U)$ be another abelian pearl, and $g,h\colon (H,T)\to(K,U)$ be sharp morphisms such that $gf = hf$. Since $f$ is surjective, we can conclude that $g=h$ as group homomorphisms, hence as pearl (sharp) homomorphisms.
Conversely, suppose that $f$ is not surjective. Let $K = H\times
(H/f(G))$, and let $U=\{(f(s),0+f(G))\mid s\in S\}$. Note that since
$f$ is a (sharp) pearl morphism, $f(s)\in T$ for all $s$, so
$(0,0)\notin U$. Now let $g,h\colon H\to K$ be defined as follows:
\begin{align*}
g(x) &= (x,x+f(G)),\\
h(x) &= (x,0+f(G)).
\end{align*}
Note that $g,h\colon (H,T)\to (K,U)$ are sharp morphisms: for all
$t\in T$ there exists $s\in S$ such that $f(s)=t$ (because $f$ is
sharp), hence $t\in f(G)$; thus, $(t,t+f(G)) = (t,0+f(G)) =
(f(s),0+f(G))\in U$; i.e., $g(T)\subseteq U$ and $h(T)\subseteq
U$. Moreover, given $(f(s),0+f(G))\in U$, we have $f(s)\in T$ so
$(f(s),0+f(G)) = g(f(s))=h(f(s))$, so $g(T)=U$ and $h(T)=U$. Thus,
both are sharp.
Since $f$ is not surjective, there exists $x\in H$ such that $x\notin
f(G)$. Therefore, $g(x) = (x,x+f(G))\neq (x,0+f(G)) = h(x)$. Hence,
$g\neq h$. However, for all $a\in G$, $g(f(a)) = (f(a),f(a)+f(G)) =
(f(a),0+f(G)) = h(f(a))$, so $g\circ f = h\circ f$. Thus, $f$ is not
an epimorphism. $\Box$
However, free and free-abelian pearls are still not projective in the corresponding sharp subcategories. In fact, once again there are no projectives at all. It is clear that if the underlying group homomorphism is onto then the pearl homomorphism is an epimorphism (even in the sharp category), though for nonabelian pearls there may be other epimorphisms.
Let $(H,T)$ be any pearl with $T$ finite, and take the identity map $i\colon (H,T)\to (H,T)$. Now let $x$ be an element not in $H$ of infinite order, and let $G=H*\langle x\rangle$ be the free product of $H$ with the infinite cyclic group; in the abelian case, take $G=H\times \langle x\rangle$. Let $S=T\cup\{x\}$, and let $s\in S$ be arbitrary. Let $f\colon (G,S)\to (H,T)$ be the map induced by the group homomorphism that maps $H$ to itself via the identity, and sends $x$ to $s$. The universal property of the free product (resp. the direct product/direct sum) guarantee such a homomorphism exists. This is a pearl homomorphism, and is sharp, since $T\subseteq f(T)\subseteq f(S)\subseteq T$. The underlying group homomorphism is onto, so the pearl map is an epimorphism. However, there are no sharp pearl homomorphisms at all from $(H,T)$ to $(G,S)$ (let alone one that factors $i$ through $f$), because the cardinality of $S$ is strictly larger than that of $T$. Thus, $(H,T)$ is not projective.
If $T$ is infinite, then a similar construction will do by replacing $\langle x\rangle$ with a free group $F$ (resp. free abelian group) of rank $\kappa$, where $\kappa\gt |T|$, replacing $S=T\cup\{x\}$ with the (disjoint) union of $T$ and the free generating set for $F$, and mapping all elements of the free generating set to elements of $T$ arbitrarily.
