Here's the proof for the decidability when the family $\mathcal{F}$ consists of all abelian simple groups $C_p$ along with all $PSL(2,q)$ for $q$ prime power.

The first remark is that a group has a quotient in $\mathcal{F}$ iff it has a nontrivial representation in $SL(2,q)$ for some $q$. (Including abelian simple groups in $\mathcal{F}$ allows me not to care about the image.)

The second remark is that a finitely generated group has a nontrivial representation in $SL(2,q)$ for some $q$ iff it has a nontrivial representation in $SL(2,K)$ for some field $K$. This does not use any kind of strong approximation, but simple facts that are basically the same as those showing that a f.g. linear group is residually finite. Namely if there's a nontrivial rep. in $SL_2(K)$, there's a nontrivial rep. in $SL_2(A)$ for some f.g. domain $A$, and since $A$ is residually a finite field we get a nontrivial rep. in $SL_2(F)$ for some finite field $F$.

Now the set of representations of a finitely presented group $\Gamma$ into $SL_2(A)$ can be described as $V(A)$, where $V$ is a $\mathbf{Z}$-defined affine variety, whose equations are explicitly defined according to a presentation of $\Gamma$. Here $V(A)=Hom(B,A)$, where $B$ is a finitely generated commutative ring, defined by a finite presentation explicitely output from the presentation of $\Gamma$.

Thus we are reduced to: given a finitely generated commutative ring $B$, determine whether there exists a field $K$ such that $Hom(B,K)$ is not reduced to a singleton $\{f_0\}$. Here $B$ comes with an explicit ring homomorphism $f_0$ onto $\mathbf{Z}$, coming from the trivial representation. Again by the standard argument, the above field $K$ can be chosen finite. Hence if we check all finite fields, we can detect when $Hom(B,K)\neq\{f_0\}$ for some field $K$. Conversely, a simple lemma (*) shows that if $Hom(B,K)=\{f_0\}$ for every field $K$, then the kernel of $f_0$ consists of nilpotent elements. Since the kernel of $f_0$ has explicit generators (the $x_i-f_0(x_i)$ where $x_i$ range over generators of $B$), in case these generators are nilpotent this can be detected in finite time. So we have an algorithm.

[Edit: proof of lemma (*): if $f_0(x)=0$ and $x$ is not nilpotent, since $B$ is a Jacobson ring there exists a quotient field $K$ in which $x$ has a nonzero image; then $K$ is finite, say of characteristic $p$ and using $f_0$, $B$ admits $K\times\mathbf{Z}/p\mathbf{Z}$ as a quotient, and thus admits two distinct homomorphisms to $K$, a contradiction.]

The argument, for each given $n$, extends to the family of all simple subquotients of $SL(n,q)$ for all prime powers $q$. Also it probably adapts, for given $(n,p)$ to the family of all simple subquotients of $SL(n,p^k)$. Now if one wishes to hold the family of say all $PSL(n,q)$ ($n$ fixed) and not its subquotients, one needs additional arguments.