This is not an answer, while I believe it is more suitable to write here instead of posting a related question. Let me know if it is not acceptable.

Set for $t\ge 0$

$$M(t):=\int_0^t(1+m(s))^2ds.$$

Then $M:\mathbb R_+\to\mathbb R_+$ is continuous (even differentiable) and strictly increasing. Define $q: \Omega\to\mathbb R_+$ by $q(t,x):=p\big(M(t), x+M(t)\big)$, where $\Omega:=\{(t,x)\in\mathbb R^2: t\ge 0, x\ge -M(t)\}$. A straightforward computation yields

\begin{eqnarray}
\partial_t q &=& \partial^2_{xx} q,\quad \forall t>0, x>-M(t),~~~~~~~~~~~~(1) \\
M(t) &=& \int_0^t (1+m(s))^2ds,\quad \forall t\ge 0~~~~~~~~~~~~ ~(2) \\
q(0,x)&=&\rho(x),\quad \forall x\ge 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (3)\\
q(t,-M(t))&=& 0,\quad \forall t>0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (4)\\
\partial_x q\big(t,-M(t)\big)&=&-m'(t)\big(1+m(t)\big)^2, \quad \forall t>0~~~~~~~~~~~~(5)
\end{eqnarray}

Here $(1-5)$ is quit similar to the system arising in Lemma 2.1 of Classical Solutions for a nonlinear Fokker-Planck equation arising in Computational Neuroscience. So my question can be reformulated as whether $(1-5)$ is well posed?

PS : (5) follows from the integration of $p$ on $\mathbb R_+$ with respect to $x$:

\begin{eqnarray}
m'(t)=\partial_t\left(\int_0^{\infty}p(t,x)dx\right)& =&\int_0^{\infty}\partial_tp(t,x)dx \\
&=& \frac{1}{(1+m(t))^2}\int_0^{\infty}\partial^2_{xx}p(t,x)dx - \int_0^{\infty}\partial_{x}p(t,x)dx \\
&=& \frac{1}{(1+m(t))^2}\partial_{x}p(t,x)|_0^{\infty} \\
&=& -\frac{1}{(1+m(t))^2}\partial_{x}p(t,0).
\end{eqnarray}