Let $y:=y_t:=\Phi_t(x)$, $x\in B\setminus\{0\}$, $\dot{y}:=d\Phi_t(x)/dt=V(y)$, $c:=C>0$. Then your condition $(V(x)\cdot x)\ge C\|V(x)\|\,\|x\|$ means that the radial speed $v_{rad}:=v_{rad}(t):=\dot{y}\cdot y/\|y\|$ is no less than $c$ times the full speed $v:=\|\dot{y}\|$. In particular, it follows that $v_{rad}>0$ while $r:=r_t:=\|y_t\|<1$, so that $r_t$ is increasing.
Suppose that $h:=r_{\infty-}<1$. Then for some real $t_0>0$ and all real $t\ge t_0$ we have $0<h/2\le r_t\le h<1$. Since $V$ is continuous and nonzero on the closed annulus between the spheres of radii $h/2$ and $h$ centered at the origin, it follows that $\|\dot{y_t}\|=\|V(y_t)\|\ge u$ and hence $v_{rad}(t)\ge ca>0$ for some real $u>0$ and all real $t\ge t_0$. This implies that for large enough $t$ we will have $r_t>1$, which contradicts the assumption $r_{\infty-}<1$.
So, $r_{\infty-}\ge1$. If $r_{\infty-}>1$, then the moving particle will hit the boundary of $B$ in a finite time.
It remains to consider the case $r_{\infty-}=1$. If $r_s=1-\delta$ for some real $s>0$ and $\delta\in(0,1)$, then the condition $v_{rad}\ge cv$ implies that, for all real $t\ge s$, the angle between the vectors $y_t$ and $y_s$ will be no greater than $\frac\delta{c(1-\delta)}$. Thus, the limit $y_{\infty-}$ will exist and be on the boundary of $B$.
So far, in addition to the condition $(V(x)\cdot x)\ge C\|V(x)\|\,\|x\|$, we have only used the condition that $V$ is nonzero and continuous away from the origin and the boundary of $B$. If we also use the Lipschitz condition, we can say a bit more: that then $r_{\infty-}=1$. (The condition that $V(0)=0$ is nowhere needed.) Indeed, assuming that $r_{\infty-}\ne1$, we have $r_{\infty-}>1$, by what has been proved. That is, $r_T=1$ for some real $T>0$. On the other hand, for some real Lipschitz constant $K>0$, all $t\in[0,T]$, $y=y_t$, and $y_*:=y/\|y\|$, $$\frac{dr}{dt}=v_{rad}\le v=\|V(y)\|\le\|V(y_*)\|+\|V(y)-V(y_*)\|\le 0+K\|y-y_*\|=K(1-r), $$ whence $$\infty=\int_{r_0}^1\frac{dr}{1-r}\le KT<\infty, $$ which is the sought contradiction.