Converse of halting problem
Here's an artificial counterexample:

Define an apparatus "aTM" like a TM, except that any input word which does not start with the symbol $a$ is immediately rejected. For words which do start with $a$, the TM is run as usual (and can accept/reject as usual).

This is clearly not Turing complete since it can't recognize the language $\{b\}$, which is decidable by a TM. Still, the halting problem on aTMs is as hard as the halting problem  for TMs.