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 $B$.