I just realized this is indeed, as Neil Strickland and Tom Goodwillie predicted, not hard, thanks to the fact that a directed union of simple groups is simple. Since homology commutes with direct limits, acyclic groups are also closed under directed unions.

So start with a group $G = G_0$ of sufficiently large cardinality. Embed it in a simple group $G_1$. Then use the Kan-Thurston result to embed $G_1$ in an acyclic group $G_2$. Repeat, obtaining a chain $G_0 \subseteq G_1 \subseteq G_2 \subseteq \dots$. The union $G_\infty$ is simple, since it's the union of the $G_{2i+1}$'s, and acyclic, since it's the union of the $G_{2i}$'s.

Thus every group $G$ embeds in a group $G_\infty$ which is simple and acyclic. In particular, there are simple acyclic groups of arbitrarily large cardinality, and the answers to all the above questions are affirmative.