Let $d\in\mathbb N$, $B_r$ denote the open ball around $0\in\mathbb R^d$ with radius $r>0$, $R>0$, $\Omega:=B_R$ and $u_0\in C_c^\infty(\Omega)$ be rotational invariant$^1$ with $u_0\not\equiv0$ and $u_0'(r)\le0$ for all $r\in(0,R)$.
We can show that the solution $u$ of \begin{equation}\left\{\begin{split}u_t&=\Delta u&\text{ in }(0,\infty)\times\Omega;\\u&=0&\text{ on }(0,\infty)\times\partial\Omega;\\u(0,\;\cdot\;)&=u_0&\text{ in }\Omega.\end{split}\right.\tag1\end{equation} is rotational invariant as well.
Moreover, we can show that $$\frac{\partial u}{\partial r}(t,r)\le0\;\;\;\text{for all }(t,r)\in[0,\infty)\times[0,R]\tag2$$ and that there are $t_0,c_0>0$ with $$\frac{\partial u}{\partial r}(t_0,R)\le-c_0\tag3.$$
How can we conclude that there isa $c>0$ such that $$u(t_0,x)\ge c\operatorname{dist}(x,\partial\Omega)?\tag4$$
Using the fundamental theorem of calculus, I only obtain $$u(t_0,x)=u(t_0,\|x\|)=u_0(0)+\int_0^{\|x\|}\frac{\partial u}{\partial r}(t_0,r)\:{\rm d}r\le u_0(x)-c_0\|x\|\tag5$$ for all $x\in\overline\Omega$, which doesn't seem to be helpful.
i.e. (abusing notation) $u_0(x)=u_0(\|x\|)$ for all $x\in\overline{B_r}$.
Since $u_r(t_0,\cdot)$ is countinuous in $\bar \Omega$ there exist $\varepsilon>0$ and $C_1>0$ s.t. $u_r(t_0,r)\le -C_1<0$ in the closure of $\Omega_1=B_R\backslash B_{R-\varepsilon}$. Therefore $$ u(t_0,r)=-\int_{R-r}^Ru_r(t_0,y)\,dy\ge C_1(R-r)=C_1\operatorname{dist}(x,\partial\Omega)\quad \hbox{in $\bar \Omega_1$.} $$ As for $\Omega_2=\Omega\backslash\bar\Omega_1$, it follows from (2) that $$ u(t_0,r)> u(t_0,R-\varepsilon)\ge C_1 \varepsilon\ge \frac{C_1 \varepsilon}{R} \operatorname{dist}(x,\partial\Omega). $$
Btw if to change the condition of rotational invariance of $u_0$ on non negativity, the statement will hold for all $t>0$ (with constant $c$ depending on $t$).