A useful general strategy to tackle such questions is to use small cancellation theory. For instance, in *Small cancellation in acylindrically hyperbolic groups*, Michael Hull proved the following statement (known before for hyperbolic and relatively hyperbolic groups):

**Theorem.** *Two finitely generated acylindrically hyperbolic groups admit a common acylindrically hyperbolic quotient.*

As a consequence, because there exist many perfect finitely generated acylindrically hyperbolic groups and that quotients of perfect groups are perfect, every finitely generated acylindrically hyperbolic group admits "many" perfect quotients. This applies in particular to braid groups: even though $B_n$ is not acylindrically hyperbolic, its quotient $B_n/Z(B_n)$ is acylindrically hyperbolic for every $n \geq 3$. 

The basic idea is the following. You start with two finitely generated acylindrically hyperbolic groups, say $A$ and $B$. Fix two finite generating sets $\{a_1, \dots, a_n\} \subset A$ and $\{b_1, \ldots, b_m \} \subset B$. Given some words $u_1, \ldots, u_n$ and $v_1, \ldots, v_m$ written respectively over $\{1, \ldots, b_m\}$ and $\{a_1, \ldots, a_n\}$, the group
$$Q:= \left( A \ast B \right) / \langle \langle a_1=u_1, \ldots, a_n=u_n, b_1=v_1, \ldots, b_m=v_m \rangle \rangle$$
is clearly a common quotient of $A$ and $B$. The difficulty is to prove something about $Q$, and it's where small cancellation is useful. If the $u_i$ and $v_j$ are chosen "sufficiently complicated", then some information can be obtained about $Q$. Of course, there is a lot a freedom in the construction (i.e. in the choice of the $u_i$ and $v_j$), which explains why "many" quotients can be constructed using this strategy.

For instance, $Q$ can be constructed so that all its finite subgroups are subgroups of $A$ or $B$. So given your favorite acylindrically hyperbolic group $G$ (say $B_n/Z(B_n)$ for some $n \geq 3$), a finite group $F$, and a torsion-free perfect acylindrically hyperbolic group $H$ (say Higman's group), then $G$, $F \ast F \ast F$, and $H$ have a common quotient, which is necessarily perfect, and all of whose finite subgroups are isomorphic to subgroups of $G$ or of $F$. Thus, if $G$ contains only finitely many isomorphism classes of finite subgroups (like $B_n/Z(B_n)$), then you obtain infinitely many perfect quotients up to isomorphism. 

In fact, the theorem above has been generalised to infinite countable collections of acylindrically hyperbolic groups by Minasyan and Osin. The construction relies on the same idea: you start with the free product of your groups, $G_1 \ast G_2 \ast \dots$, and you add complicated relations (i.e. satisfying some good small cancellation condition) in order to merge each $G_i$ into each $G_j$. Of course, you have to add infinitely many relations, which makes the quotient infinitely presented, but this also means that you have much more freedom in the construction. In particular, it is possible to create uncountably many perfect quotients up to isomorphism.

However, in such constructions, the quotients are always infinite. If you are looking for finite quotients, then the strategy is not relevant.