Are hash functions really quantum resistant for commitment schemes?
No.  I can see how that sounds like it could work, but it doesn't actually work.  You're missing an important detail about how we implement classical functions on a quantum computer.  The usual way to implement non-reversible functions with reversible logic is to use ancilla bits (i.e., constant inputs, and junk outputs that are discarded).

For instance, suppose you want to implement the function $f(x,y) = x \land y$ (it computes the boolean AND of its two inputs).  You can make this invertible by instead implementing $g(x,y,z) = ((x \land y) \oplus z, x, y)$ (here $\oplus$ denotes XOR).  Here $g$ is called a Toffoli gate.  Note that $g$ is invertible, even though $f$ is not.  Moreover, we can compute $f$ by noting that $g(x,y,0) = (f(x,y,0),x,y)$, so if you want to compute $f$ in a reversible way, you feed in $x$, $y$, and a constant $0$ to $g$, then discard all but the first bit of output.  Just because you can invert $g$ doesn't mean you can invert $f$.  First, given only $f(x,y)$, we don't know what value to place into the other output bits of $g$.  Second, if we made random guesses at those outputs and then inverted $g$, that won't necessarily give us an inverse of $f$ (try it!).

The same is true for SHA2.  We know how to implement SHA2 on a quantum computer.  It uses ancilla bits.  Those ancilla bits prevent you from inverting SHA2 by trying to reverse the quantum computation.