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.
Update.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.