This is a straightforward comparison argument. Let $\omega_n$ be the volume of $\partial B_1\subset \mathbb{R}^{n+1}$.

For generic $R$, one has $\partial B_R \cap T=\tau$ a smooth submanifold. By Alexander duality, there is a subset, $\Omega$, of $ \partial B_R\setminus \tau$ so $\partial \Omega=\tau$. Clearly, $\mathcal{H}^n(\Omega)\leq \omega_n R^n$. In fact, up to replacing $\Omega$ by it's complement one has $\mathcal{H}^n(\Omega)\leq \frac{1}{2}\omega_n R^n$.

By the area minimization property, $$\mathcal{H}^n(\Sigma\cap B_R)\leq \mathcal{H}^n(\Omega)\leq \omega_n R^n.$$

The monotonicity formula ensures the bound holds for all $R$.

In fact, this argument should work (using slicing) for an area minimizing integral current.

This is a straightforward comparison argument. Let $\omega_n$ be the volume of $\partial B_1\subset \mathbb{R}^{n+1}$.

For generic $R$, one has $\partial B_R \cap T=\tau$ a smooth submanifold. By Alexander duality, there is a subset, $\Omega$, of $ \partial B_R\setminus \tau$ so $\partial \Omega=\tau$. Clearly, $\mathcal{H}^n(\Omega)\leq \omega_n R^n$. In fact, up to replacing $\Omega$ by it's complement one has $\mathcal{H}^n(\Omega)\leq \frac{1}{2}\omega_n R^n$.

By the area minimization property, $$\mathcal{H}^n(\Sigma\cap B_R)\leq \mathcal{H}^n(\Omega)\leq \omega_n R^n.$$

The monotonicity formula ensures the bound holds for all $R$.

In fact, this argument should work (using slicing) for an area minimizing integral $\mathbb{Z}_2$ current.