if $L\in NP\cap Co-NP$ is NP-Hard, then $NP=Co-NP$
Suppose $L$ is an NP-hard problem in coNP, and let $M$ be any problem in NP. Since $L$ is NP-hard, there is a polytime reduction $f$ such that $x \in M$ iff $f(x) \in L$. Since $L$ is in coNP, this gives a coNP algorithm for $M$: given an input $x$, compute $f(x)$ and apply the coNP algorithm for $L$. This shows that $M$ is in coNP. In other words, NP is a subset of coNP. The same argument also shows the reverse inclusion, and so NP equals coNP.

The key idea is to avoid Cook reductions, for the reasons you mention.