There does not exist any function $p:F\mapsto p_F$ as in the question, to prove it by contradiction suppose such a function $p$ exists. In the following let $k+A=\{k+a;a\in A\}$ for any $k\in\mathbb{R}^2$, $A\subseteq\mathbb{R}^2$
To see why, consider the triangle $T_0$ with vertices $(-1,0),(1,0)$ and $(0,1)$. Let $r:(x,y)\mapsto(x,-y)$ be the reflection respect to the $x$ axis.
Now consider for each natural $n$ the triangle $T_n=r^n(T_0+10(n,0))$, and let $p_{T_n}=(x_n,y_n)$. Note that $d_H(T_n,T_{n+1})=10$, so $\sqrt{(x_{n+1}-x_n)^2+(y_{n+1}-y_n)^2}=d(p_{T_n},p_{T_{n+1}})\leq 10$. This, of course, implies that $x_{n+1}-x_n\leq10$. Moreover, $(y_{n+1}-y_n)^2=(|y_{n+1}|+|y_n|)^2$, since $y_n$ and $y_{n+1}$ have opposite signs.
This will allow us to deduce by contradiction that $y_n\to 0$: if not, we would have some $\varepsilon>0$ and infinitely many $n$ such that $|y_n|>\varepsilon$, so $x_{n+1}-x_n<\sqrt{100-\varepsilon^2}$. This implies that for big enough $n$, we will have $x_n-x_0<10n-2$, which is impossible since $10n-2=d(T_n,T_0)$.
Similarly, let $S_0$ be the triangle with vertices $(-1,1),(1,1)$ and $(0,0)$, let $s$ be the reflection around the line $y=1$ and let $S_n=s^n(S_0+10(n,0))$. Then letting $p_{S_n}=(z_n,w_n)$, we can prove that $w_n\to 1$ as before. So for big enough even $n$ we have $d_H(T_n,S_n)=\frac{\sqrt{2}}{2}$ and $d(p_n,q_n)\geq0.9$, the contradiction we were looking for.