Lemma: $H$ is the reflection of $O$ over the angle bisector of $\angle BAC$ (henceforth 'the' reflection)
Proof: Let $H'$ be the reflection of $O$, and let $B'$ be the reflection of $B$.
Then reflection takes $\angle ABH'$ to $\angle AB'O$.
$\Delta ABB'$ is equilateral, and $O$ lies on the perpendicular bisector of $\overline{AB}$
It's well known that $O$ lies strictly inside $\Delta ABC$ (since it's acute), meaning that $\angle ABH' = \angle AB'O = 30^{\circ},$ from which it follows that $\overline{BH'} \perp \overline{AC}$ . Similarly, $\overline{CH'} \perp \overline{AB}$. Since $H'$ lies on two altitudes, $H'$ is the orthocenter, as desired.
End Lemma
So $\overline{OH}$ is perpendicular to the angle bisector of $\angle OAH$, which is the same line as the angle bisector of $\angle BAC$, meaning that $\Delta APQ$ is equilateral.
Let its side length be $s$, and let $PH=t$, where $0 < t < s, t \neq s/2$ because $O$ lies strictly within $\angle BAC$, as must $H$, the reflection of $O$. Also, it's easy to show that if $O=H$ in a general triangle, it's equilateral, and we know $\Delta ABC$ is not equilateral. Hence H is not on the bisector of $\angle BAC \implies t \neq s/2$. Let $\overrightarrow{BH}$ intersect $\overline{AC}$ at $P_B$.
Since $\Delta HP_BQ$ and $BP_BA$ are 30-60-90 triangles, $AB=2AP_B=2(s-QP_B)=2(s-HQ/2)=2s-HQ=2s-(s-t)=s+t$
Similarly, $AC=2s-t$
The ratio $\frac{[APQ]}{[ABC]-[APQ]}$ is $\frac{AP \cdot AQ}{AB \cdot AC - AP \cdot AQ} = \frac{s^2}{(s+t)(2s-t)-s^2}$
The denominator equals $(1.5s)^2-(.5s-t)^2-s^2$ where $.5s-t$ can equal any value in $(-.5s, .5s)$ except $0$. Therefore, the denominator can equal any value in $(s^2, 5s^2/4)$, and the ratio is any value in $\boxed{\left(\frac{4}{5},1\right)}.$
Note: It's easy to show that for any point $H$ on $\overline{PQ}$ except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.