Here is a very, very rough sketch of a possible proof...
Let $\bar{\rho} (\epsilon) := \rho + \epsilon \rho'$. The limit can then be written in the form
$\displaystyle\lim_{\epsilon \to 0} \frac{1}{\epsilon}\left[ F (\bar{\rho} (\epsilon) + \epsilon^2 \rho'') - F (\bar{\rho} (\epsilon))\right]$
The "Taylor expansion" of $F$ is the following
$F (\bar{\rho} (\epsilon) + \epsilon^2 \rho'') = F (\bar{\rho} (\epsilon)) + \langle \nabla F (\bar{\rho} (\epsilon)), \epsilon^2 \rho''\rangle + \omicron (\epsilon^4)$
where $\nabla F (\bar{\rho} (\epsilon))$ is the "functional gradient" of $F$. Then, we have that
$\displaystyle\lim_{\epsilon \to 0} \frac{1}{\epsilon}\left[ F (\bar{\rho} (\epsilon) + \epsilon^2 \rho'') - F (\bar{\rho} (\epsilon))\right] = \lim_{\epsilon \to 0} \left[\frac{1}{\epsilon} \langle \nabla F (\bar{\rho} (\epsilon)), \epsilon^2 \rho''\rangle + \omicron (\epsilon^3)\right]$
If we can show that
$\frac{1}{\epsilon} \langle \nabla F (\bar{\rho} (\epsilon)), \epsilon^2 \rho''\rangle = \langle \nabla F (\bar{\rho} (\epsilon)), \epsilon \rho''\rangle$
then the limit becomes
$\displaystyle\lim_{\epsilon \to 0} \frac{1}{\epsilon}\left[ F (\bar{\rho} (\epsilon) + \epsilon^2 \rho'') - F (\bar{\rho} (\epsilon))\right] = \lim_{\epsilon \to 0} \left[\langle \nabla F (\bar{\rho} (\epsilon)), \epsilon \rho''\rangle + \omicron (\epsilon^3)\right] = 0$