Choose $x_{0}\in\partial B(p,R)$, so that $d(x_{0},p)=R$, and denote by $\rho(x)=d(x,x_{0})$. The Laplacian comparison theorem gives $\Delta\rho^{2}\leq 4n$ in the sense of distributions. Let $\varphi(x)=\psi(\rho(x))$ where

$$\psi(t)=\left\{\begin{array}[]{lll}1&\mbox{if $0\leq t\leq R-1$},\\ \frac{1}{2}(R+1-t)&\mbox{if $R-1<t<R+1$},\\ 0&\mbox{if $t\geq R+1$}.\end{array}\right.$$

Then $\varphi$ is a nonnegative Lipschitz function supported in $B(x_{0},R+1)$, and we have that

$$\begin{split}\int_{M}\varphi\Delta\rho^{2}dV_{g}&=-\int_{B(x_{0},R+1)}\nabla\varphi\cdot\nabla\rho^{2}dV_{g}=-2\int_{B(x_{0},R+1)}\rho|\nabla\rho|^{2}\psi^{\prime}(\rho(x))dV_{g}\\ &=\int_{B(x_{0},R+1)\backslash B(x_{0},R-1)}\rho dV_{g}\\ &\geq(R-1)\mathrm{Vol}(B(x_{0},R+1)\backslash B(x_{0},R-1)),\end{split}$$

and also

$$\int_{M}\varphi\Delta\rho^{2}dV_{g}\leq 4n\int_{B(x_{0},R+1)}\varphi dV_{g}\leq 4n\mathrm{Vol}(B(x_{0},R+1)).$$

Notice that $B(p,1)\subset B(x_{0},R+1)\backslash B(x_{0},R-1)$ and so the previous two equations give

$$(R-1)\mathrm{Vol}(B(p,1))\leq 4n\mathrm{Vol}(B(x_{0},R+1)).$$

The conclusion follows from the fact that $B(x_{0},R+1)\subset B(p,2(R+1))$. ∎