The answer is YES.

Let $b\colon M\to \mathbb R$ be a Busemann function for a ray $\gamma$ from $p$, so that $b(p)=0$ and $b(x)\le 0$ for any $x\in \gamma$.

Set $$L_t=b^{-1}(t)\ \ \text{and}\ \ L^-_t=b^{-1}(-\infty,t].$$ Note that the sublevel sets $L_t^-$ are mean curvature concave for all $t$. In particular, any area minimizing hypersurface in $L_t^-$ with the boundary in $L_t$ lies in $L_t$.

Fix $t<0$. From above, $$\mathop{\rm area}\partial B(p,r)\ge\mathop{\rm area}(\partial B(p,r)\cap L_t^-)\ge \mathop{\rm area}( B(p,r)\cap L_t)\ge \mathop{\rm area}( B(p,R)\cap L_t);$$ i.e., the inequality holds for $c(p,R)=\mathop{\rm area}( B(p,R)\cap L_t)$.

It remains to choose $t$ and $R$ so that $\mathop{\rm area}( B(p,R)\cap L_t)>0$; $R=2$ and $t=1$ will do the job.

The answer is YES.

Let $b\colon M\to \mathbb R$ be a Busemann function for a ray $\gamma$ from $p$, so that $b(p)=0$ and $b(x)\le 0$ for any $x\in \gamma$.

Set $$L_t=b^{-1}(t)\ \ \text{and}\ \ L^-_t=b^{-1}(-\infty,t].$$ Note that the sublevel sets $L_t^-$ are mean curvature concave for all $t$. In particular, any area minimizing hypersurface in $L_t^-$ with the boundary in $L_t$ lies in $L_t$.

Fix $t<0$. From above, $$\mathop{\rm area}\partial B(p,r)\ge\mathop{\rm area}(\partial B(p,r)\cap L_t^-)\ge \mathop{\rm area}( B(p,r)\cap L_t)\ge \mathop{\rm area}( B(p,R)\cap L_t);$$ i.e., the inequality holds for $c(p,R)=\mathop{\rm area}( B(p,R)\cap L_t)$.

It remains to choose $t$ and $R$ so that $\mathop{\rm area}( B(p,R)\cap L_t)>0$; $R=2$ and $t=-1$ will do the job.