When $n = 2$ the sort of examples you require does not exist. This is due to Fischer-Colbrie and Schoen, "The structure of complete stable minimal surfaces in 3-manifolds of non-negative scalar curvature".

https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160330206

Essentially: requiring just that the minimal surface is area minimizing for compact perturbations up to second order (which is weaker than the strict area-minimizing condition you asked for), they prove (among other things) that the only "stable" minimal surface in $\mathbb{R}^3$ is the plane.

The reason is basically that the second variation of the mean curvature gives a Laplace equation with a potential, which in $n = 2$ can be related to the scalar curvature of the minimal surface. And non-trivial minimal surfaces in $\mathbb{R}^3$ all have negative scalar curvature.

When $n = 2$ the sort of examples you require does not exist. This is due to Fischer-Colbrie and Schoen, "The structure of complete stable minimal surfaces in 3-manifolds of non-negative scalar curvature".

https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160330206

Essentially: requiring just that the minimal surface is area minimizing for compact perturbations up to second order (which is weaker than the strict area-minimizing condition you asked for), they prove (among other things) that the only "stable" minimal surface in $\mathbb{R}^3$ is the plane.

The reason is basically that the second variation of the area gives a Laplace equation with a potential, which in $n = 2$ can be related to the scalar curvature of the minimal surface. And non-trivial minimal surfaces in $\mathbb{R}^3$ all have negative scalar curvature.

The area of the helicoid grows like $R^3$, if I am not mistaken.

When $n = 2$ the sort of examples you require does not exist. This is due to Fischer-Colbrie and Schoen, "The structure of complete stable minimal surfaces in 3-manifolds of non-negative scalar curvature".

https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160330206

Essentially: requiring just that the minimal surface is area minimizing for compact perturbations up to second order (which is weaker than the strict area-minimizing condition you asked for), they prove (among other things) that the only "stable" minimal surface in $\mathbb{R}^3$ is the plane.

The reason is basically that the second variation of the mean curvature gives a Laplace equation with a potential, which in $n = 2$ can be related to the scalar curvature of the minimal surface. And non-trivial minimal surfaces in $\mathbb{R}^3$ all have negative scalar curvature.