Suppose $({\cal E,M})$ is a factorization system on $\cal C$ such that $\cal M$ consists of monomorphisms and $F$ preserves $\cal E$-morphisms. If $X$ is an $F$-algebra, define an $\cal M$-subalgebra of $X$ to be a morphism $Y\to X$ of $F$-algebras that is an $\cal M$-morphism. Then if $A$ is the initial $F$-algebra, the map $A\to X$ lies in $\cal E$ if and only if $X$ has no proper $\cal M$-subalgebras.
In the "if" direction, note that the assumption that $F$ preserves $\cal E$-morphisms implies that $({\cal E,M})$-factorizations lift to $F$-algebras. In particular, if we factor the unique $F$-algebra morphism $A\to X$ as $A\to Z \to X$, then the unique lifting property of $F A \to F Z$ (which is in $\cal E$) against $Z\to X$ (which is in $\cal M$) makes $Z$ an $F$-algebra and the two maps $F$-algebra maps. In particular, $Z$ is an $\cal M$-subalgebra of $X$. Thus, if $X$ has no proper $\cal M$-subalgebras, then $Z\cong X$ and hence $A\to X$ is in $\cal E$.
Conversely, if $A\to X$ is in $\cal E$, suppose $Z\to X$ is an $\cal M$-subalgebra. Then $A\to X$ factors through $Z$ (since $A$ is the initial $F$-algebra), but then $Z\to X$ must be an isomorphism since it is an $\cal M$-morphism that is factored through by some $\cal E$-morphism.
In particular, if $\cal C$ admits (epi, strong mono) factorizations, then $A\to X$ is epi if and only if $X$ has no proper strong-subalgebras.
EDIT: Note that if $X$ has no proper $\cal M$-subalgebras, then $\xi : F X \to X$ is in $\cal E$. For we can $({\cal E,M})$-factor it as $F X \to Z\to X$ and show that $Z$ becomes an $\cal M$-subalgebra. The assumption on $X$ then implies $Z\cong X$, hence $\xi\in\cal E$.
On the other hand, as Valery pointed out, taking $F=\rm Id$ shows that the converse can't hold. Any endo-epimorphism $X\to X$ (e.g. $(+1) : \mathbb{Z}\to\mathbb{Z}$) gives an $F$-algebra $\xi$ that is an epimorphism, but $0 = A \to X$ will rarely be epi.