Maybe I'm missing something, but let $f(H)=\mathrm{tr}(DHAH)$, then

$$f(\alpha H)=\alpha^2f(H)$$

for every real $\alpha$. Obviously $f(0)=0$. So the problem boils down to find an $H$ such that $f(H)>0$ and another one such that $f(H)<0$.

Now, denoting by $h_k$ the $k-$th column of $H$,

$$(HAH)_{ij}=(h_i)^tAh_j$$

Therefore $f(H)=\sum d_i (h_i)^tAh_i$, if $D=\mathrm{diag}(d_1,\ldots, d_n)$.

If, wlog, $d_1>0$ and $d_2<0$, then we are obviously done.

If every $d_i$ has the same sign, but $A$ has a positive and a negative eigenvalue, then again we are done.

If every $d_i$ is wlog positive (or null) and $A$ is semi-definite, then $f(H)\geq0$ for every $H$ symmetric and $f(S)\leq 0$ for every $S$ skew-symmetric, so, given $H$ such that $f(H)\neq0$, we can always find $\alpha$ such that $f(\alpha H)=-f(S)$.