How secure is a hash based signature scheme after signing assuming quantum computers?
Two things:

The question &quot;how hard is the oracle-based collision-finding problem be for a Quantum Computer&quot; is an open question; we know that it is somewhere between $O(\sqrt[3]{n})$ and $O(\sqrt{n})$ effort (I recall a paper showing that $O(\sqrt[3]{n})$ Oracle queries is the lower bound, but I forget what paper that was) - we don't know where in the range it actually is.  Brassard et al showed an algorithm that found collisions with $O(\sqrt[3]{n})$ Oracle queries; however Dan pointed out that the other costs associated with the algorithm made it actually more expensive than just running a large number of second-preimage finders in parallel [1].  What is currently unknown whether there exists another algorithm that uses $o(\sqrt{n})$ Oracle queries [2], and doesn't have the other expenses that the Brassard algorithm has...

The three current 'hash based signature standards', namely, LMS, XMSS and Sphincs+ don't actually rely on collision resistance, even when doing the initial hash of the message.  What all three do is, when the signer is given a message $M$, the signer selects an unpredictable $R$ and then computes $\text{Hash}(R + M)$ (and then uses that hash for the rest of the signature operation).  If the adversary submits a message to be signed, he doesn't know what the value $R$ that the signer will select; instead, to perform a collision attack, he would need to select two distinct messages $M, M'$ for which $\text{Hash}(R + M) = \text{Hash}(R + M')$ with nontrivial probability (given the probability distribution of $R$), and that would appear to be far more difficult than finding a collision.



[1] The Brassard algorithm is essentially a large multi-target second preimage attack, and Dan's complaint is that the quantum logic to do a recognization over a large sparse unstructured space is quite expensive. In an unpublished work, I looked at the Brassard algorithm, and searched for the 'sweet spot' between 'number of targets per searcher' and 'number of searchers' - based on the published SHA-256 Quantum circuit, and a sketched out target recognizer circuit I drew, I came to the conclusion that this made the problem circa a factor of 1000 cheaper to attack than a single preimage problem (and the sweet spot was circa a million targets per searcher).
[2] For those who are unfamiliar with little-oh notation, $o(\sqrt{n})$ means 'fewer than $c\sqrt{n}$ queries, for any $c &gt; 0$ (as long as $n$ is large enough, and the meaning of &quot;large enough&quot; depends on $c$).  More colloquially, $o(\sqrt{n})$ means 'grows slower than $\sqrt{n}$'