At a minimum, you should assume $F\ge 0$. I also think $F$ should be continuous, although it's possible to make sense of the problem for some discontinuous $F$.
Even for smooth $F$ you will not have $|\nabla T| =F$ everywhere. Imagine a radially symmetric function $F$ such that $F=1$ on most of the space, but $F$ rapidly grows toward $(0,0,0)$, presenting a kind of obstacle. Then if $S=(-1,0,0)$ and $(x,y,z)=(1,0,0)$, the optimal connecting curves will go around the obstacle. By symmetry, they will not be unique, and this will lead to $T$ not being differentiable at $(1,0,0)$.
But you can prove that $|\nabla T|=F$ at almost every point, as follows. First, $T$ is locally Lipschitz, because $T(q)- T(p)$ does not exceed the cost of moving from $p$ to $q $ in straight line. Therefore, $T$ is differentiable almost everywhere (Rademacher's theorem). Let's fix a point of differentiability, call it $p$. Consider a small ball $B_r(p)$ centered at $p$. By continuity of $F$, all values of $F$ in this ball are close to $F(p)$: $$F(p)-\epsilon \le F\le F(p)+\epsilon\tag1$$ Hence, the Lipschitz constant of $T$ on this ball does not exceed $F(p)+\epsilon$, which in the limit $r\to 0$ implies $|\nabla T(p)|\le F(p)$.
To show the reverse inequality, let $q$ be a point of $\partial B_r(p)$ at which $T$ attains its minimum on $\partial B_r(p)$. Since every path from $p$ to $S$ has to cross $\partial B_r(p)$, it follows that $$T(p)\ge (F(p)-\epsilon)r+T(q) \tag2$$ where the first term comes from traveling from $p$ to $\partial B_r(p)$. On the other hand, the definition of derivative implies $$|T(p)-T(q)|\le |\nabla T(p)|r + o(r)\tag3$$
From (2) and (3), $|\nabla T(p)|\ge F(p)$.