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, either (i) $r_T=1$ for some real $T>0$ (which should be fine for us) or (ii) $r_{\infty-}=1$ but $r_t<1$ for all real $t\ge0$.
It remains to consider the case (ii). 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 $B$.
Added: A detailed and improved (?) version of the latter paragraph (without using angles) is as follows: For real $t>0$ and $\Delta t\downarrow0$, we have $\Delta y_t:=y_{t+\Delta t}-y_t=V(y_t)\Delta t+o(\Delta t)$ and hence $\|\Delta y_t\|=\|V(y_t)\|\Delta t+o(\Delta t)\le\frac1c\,v_{rad}(t)\Delta t+o(\Delta t)$. Also, $\Delta r_t:=r_{t+\Delta t}-r_t=v_{rad}(t)\Delta t+o(\Delta t)$, so that $\|\Delta y_t\|\le\frac1c\,\Delta r_t+o(\Delta t)$. Integrating/telescoping this and using the triangle inequality for the norm $\|\cdot\|$, we see that for all $s$ and $t$ such that $0<s<t<\infty$ we have $\|y_t-y_s\|\le\frac1c\,(r_t-r_s)$. Since $r_{\infty-}=1$, we conclude that, by the Cauchy convergence criterion, the limit $y_{\infty-}$ exists and is 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 case (ii) must take place. (The condition that $V(0)=0$ is nowhere needed.) Indeed, assuming the contrary, we have case (i), 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.
If the Lipschitz condition fails to hold near the boundary of $B$, then of course case (i) may take place. E.g., if $d=1$ and $V(y)=y\sqrt{1-|y|}$, then with $y_0=x\in(-1,1)\setminus\{0\}$ we have $$y_t=\Big[1-\tanh ^2\left(\tfrac{t}{2}-\tanh ^{-1}\sqrt{1-|x|}\right)\Big]\,\text{sign} \,x $$ and $y_T=\text{sign}\,x$ for $T=2\tanh ^{-1}\sqrt{1-|x|}<\infty$.