Fix any $x\in B\setminus\{0\}$. Let $y:=y_t:=\Phi_t(x)$, $r:=r_t:=\|y_t\|$, $\dot{y}:=d\Phi_t(x)/dt=V(y)$ (the velocity), $v:=\|\dot{y}\|=\|V(y)\|$ (the speed), $c:=C>0$. Then for all $t>0$ such that $0<r_t<1$ we have \begin{equation*} \dot r=\frac{d\|y\|}{dt}=\frac{y\cdot\dot y}{\|y\|}=\frac{y\cdot V(y)}{\|y\|}\le\|V(y)\|=v; \tag{1} \end{equation*} on the other hand, your condition $(V(x)\cdot x)\ge C\|V(x)\|\,\|x\|$ implies that \begin{equation*} \dot r=\frac{y\cdot V(y)}{\|y\|}\ge c\|V(y)\|=cv>0, \tag{2} \end{equation*} so that $r_t$ is increasing in $t\in[0,T)$, where \begin{equation*} T:=\inf\{t>0\colon r_t=1\}; \end{equation*} recall that $\inf\emptyset$ is defined as $\infty$. Thus, the limit $h:=r_{T-}$ exists.
Let us show that $h=1$. Indeed, suppose the contrary. Then $T=\infty$ and there is some real $t_0>0$ such that \begin{equation*} 0<h/2\le r_t\le h<1 \text{ for all real }t\ge t_0. \tag{3} \end{equation*} Since $V$ is continuous and nonzero on the closed "annulus" $A:=\{z\in\mathbb R^n\colon h/2\le\|z\|\le h\}$, it follows that $\|V(z)\|\ge u$ for some real $u>0$ and all $z\in A$. So, for all real $t\ge t_0$ we have $\|V(y_t)\|\ge u$ and hence, by (2), $\dot r\ge cu>0$. This implies that for some real $t\ge t_0$ we will have $r_t=1$, which contradicts (3). Thus, \begin{equation*} r_{T-}=1. \tag{4} \end{equation*}
Let us now show that the Lipschitz condition on the vector field $V$ implies that, in fact, $T=\infty$. Indeed, for some real Lipschitz constant $K>0$, all $t\in[0,T)$, $y=y_t$, and $y_*:=y/\|y\|$, by (1), $$\frac{dr}{dt}=\dot r\le \|V(y)\|\le\|V(y_*)\|+\|V(y)-V(y_*)\|\le 0+K\|y-y_*\|=K(1-r), $$ whence $-\frac{d}{dt}\,\ln(1-r)\le K$. So, in view of (4), $$\infty=\lim_{t\uparrow T}(\ln(1-r_0)-\ln(1-r_t))\le KT, $$ which does imply that $T=\infty$.