Is $NP$ "minimal", i.e. does $\Pi\notin NP$ imply $\Pi$ is $NP$-hard? Suppose $\Pi$ is a decidable decision problem.


  Does $\Pi\not \in NP$ imply $\Pi$ is $NP$-Hard?


Edit: if we assume there exists $\Pi\in coNP\setminus NP$ then we are done. Can we refute the claim without any unknown assumptions?
If you assume that $\mathsf{NP}\neq\mathsf{coNP}$ then any coNP-complete problem gives a counterexample. I would guess that one can refute your conjecture unconditionally.