A short answer is "why not?". A longer answer would be to look at the known examples of non-Hopfian groups and try to make them lacunary hyperbolic. A quite general construction can be found in our paper with Dani Wise (Sapir, Mark; Wise, Daniel T.
Ascending HNN extensions of residually finite groups can be non-Hopfian and can have very few finite quotients. J. Pure Appl. Algebra 166 (2002), no. 1-2, 191–202.). See Lemma 3.1 there, in particular. It is quite possible that this construction or its slight modification can be lacunary hyperbolic.

<b> Update.</b>Another way to construct an example is the following. Start with the free group $F_2=\langle a,b\rangle$. Pick two words $U(a,b), V(a,b)$ satisfying small cancelation. That will be the non-injective surjective endomorphism. To make it non-injective, pick a word $W(x,y)$ and impose the relation $W(U, V)=1$. To make it surjective, pick two  words $P(a,b), Q(a,b)$, and impose the relations $P(U,V)=a, Q(U,V)=b$. Now to make the map $a\to U, b\to V$ an endomorphism, for every relation $S(a,b)=1$ introduced already, we need to add the relation $S(U,V)=1$, then apply the same operation to the resulting presentation, etc. This defines an infinite presentation naturally subdivided into finite subsets. It remains to choose the words $U,V,W, P, Q$ so that each finite piece of the presentation defines a hyperbolic groups and the whole presentation is lacunary hyperbolic. Some kind of small cancelation theory may help here.