I am not sure whether such a function $p(r)$ actually exists. My intuition is choosing an exponential-type function that decreases rapidly for values of $r$ away from $t$ can potentially ensure that the integral goes to $0$ as $\frac{t}{\sqrt{A}} \rightarrow \infty$. But I couldn't prove this.
One more constraint: $p(r)$ should be independent of $t$.
As stated it is not possible to find such a $p$. To see this, we pick $A=1=w$ and let $t\rightarrow \infty$ (that's all compatible with the assumptions). Then we note that the integrand is positive and hence we have for $t\geq 2$
$$I(t,1)\geq \int_1^2 \frac{t^2}{4} p(r) \exp\left( -\frac{1}{r^2}+\frac{2}{r} \right)dr \geq \frac{\exp\left(-\frac{1}{4} \right)}{4} \left(\int_1^2 p(r) dr \right) t^2. $$
The integral over $p$ needs to vanish in order to make $I(t,1)=o(1)$, which contradicts the assumption that $p>0$.