A cohomological bound can be obtained as follows.

**Proposition** The affine stratification number of $H_{g,n}$ is $1$ for $n > 0$, $0$ for $n=0$.

*Proof* Since $H_{g,0}$ is affine we are only interested in the case $n > 0$. Since $H_{g,n+1} \to H_{g,n}$ is an affine morphism for $n>0$, it's enough to do the case $n=1$. Write $H_{g,1}$ as the disjoint union of the locus where the marked point is (resp. is not) a Weierstrass point. Both these loci are affine since the projection down to $H_g$ is an affine morphism on each of them. To see that $H_{g,1}$ is not affine, note that it contains many complete curves - specifically, any fiber of $H_{g,1} \to H_g$ is proper. **QED**

This implies that not just the rational cohomology, but the cohomology of any constructible sheaf (over $\mathbb Q$, to take care of stacky issues) on $H_{g,n}$ vanishes above degree $2g+n$ for $n > 0$. Also vanishing theorems for constructible sheaves - see the paper of Roth and Vakil.

In general I would expect this bound to be sharp. By this I mean that for a given $g$, there will only be finitely many $n$ for which $H^{2g+n}(H_{g,n},\mathbb Q) = 0$. I have no idea how one would prove this, though.

Now let me prove the claim I made in a comment.

**Proposition** The space $H_g$ has the rational cohomology of a point for all $g$.

*Proof* Note that $H_g = M_{0,2g+2}/S_{2g+2}$. So it will be enough to prove that $M_{0,2g+2}/S_{2g+1}$, parametrizing $2g+1$ unordered marked points on $\mathbb P^1$ and a distinguished other point, has the rational cohomology of a point. If we put the distinguished marking at infinity, we get an isomorphism $M_{0,2g+2}/S_{2g+1} = B(\mathbb A^1,2g+1)/G$. Here $B(\mathbb A^1,n)$ denotes the configuration space of $n$ unordered marked points on $\mathbb A^1$, and $ G= \mathbb G_m \ltimes \mathbb G_a$ denotes the group of Möbius transformations fixing the point at infinity. In fact the projection from the configuration space to $M_{0,2g+2}/S_{2g+1}$ is a trivial $G$-bundle.

The rational cohomology of $B(\mathbb A^1,n)$ was computed by Arnold in his famous paper on the cohomology of the braid group: it has the rational cohomology groups of a circle. And $G$ is homotopic to a circle. So $M_{0,2g+2}/S_{2g+1}$ has the rational cohomology of a point, as claimed. **QED**

**Proposition** $H^0(H_{g,1},\mathbb Q) \cong H^2(H_{g,1},\mathbb Q) \cong \mathbb Q$ and all other cohomology groups vanish.

*Proof* Apply the Leray spectral sequence for $f:H_{g,1} \to H_g$. We have $R^0 f_\ast \mathbb Q \cong R^2 f_\ast \mathbb Q \cong \mathbb Q$. The locally constant sheaf $R^1 f_\ast \mathbb Q$ has vanishing cohomology, because every point of $H_g$ has the hyperelliptic involution as a nontrivial stabilizer, and the hyperelliptic involution acts as $-1$ on the stalk of the local system. The result follows. **QED**

The cohomologies of $H_{g,2}$ and $H_{g,3}$ have also been computed for all $g$ by Orsola Tommasi in her Ph.D. thesis.

Let me also mention two results about the cohomology of $H_{g,n}$ that I can prove but haven't written down yet.

**Proposition** The Leray spectral sequence for $H_{g,n} \to H_g$ degenerates for all $g,n$.

This is in contrast with the Leray spectral sequence for $M_{g,n} \to M_g$, which typically does not degenerate. It is really very special to families of hyperelliptic curves. It means that understanding the cohomology of $H_{g,n}$ is equivalent to understanding the cohomology of $H_g$ with coefficients in a symplectic representation.

**Proposition** There is an analogue of Harer stability for moduli of hyperelliptic curves: $H^k(H_{g,n},\mathbb Q) \cong H^k(H_{g+1,n},\mathbb Q)$ for $g \gg k$.

This isn't really "my" theorem - it follows from the work of Randal-Williams and Wahl.