I think that the phenomena is _not_ stable under perturbation of the metric, i.e., a small perturbation can cause the minimizer to be smooth.

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First, let me explain how it is not stable under perturbation of the "link." 

If you have a smooth submanifold $\Gamma^{n-1} \hookrightarrow \mathbb{S}^n \hookrightarrow \mathbb{R}^{n+1}$ so that the cone over $\Gamma$, $C_1(\Gamma)=\{tx : x\in \Gamma,t\in [0,1]\}$, is area-minimizing, then there exist arbitrarily small perturbations $\Gamma_\epsilon \hookrightarrow \mathbb{S}^{n}$ of $\Gamma$ so that the solution to the Plateau problem (i.e., the area minimizer with boundary $\Gamma_\epsilon$) for $\Gamma_\epsilon$ is regular. 

In general, this is a consequence of the <a href="http://www.ams.org/mathscinet-getitem?mr=809969">work</a> of Hardt and Simon, who showed that if $C(\Gamma)=\{tx : x\in\Gamma, t\geq 0\}$ is area minimizing (this is equivalent to $C_1(\Gamma)$ solving the Plateau problem for $\Gamma$), then $\mathbb{R}^{n+1}\setminus C(\Gamma)$ is foliated by _smooth_, area minimizing hypersurfaces which are asymptotic to $C(\Gamma)$ at infinity. For the Simons' cone you mentioned, this fact was proven earlier by Bombieri, De Giorgi, and Giusti as part of their <a href="http://www.ams.org/mathscinet-getitem?mr=250205">proof</a> that the Simons' cone is area minimizing. 

Now, because these smooth surfaces foliate $\mathbb{R}^{n+1}\setminus C(\Gamma)$, and are asymptotic to $C(\Gamma)$, we can find one, say $\Sigma_\epsilon$, which intersects $\mathbb{S}^n$ in a (smooth) surface, say $\Gamma_\epsilon$ which is arbitrarily close to $\Gamma$. Because $\Sigma_\epsilon$ is area minimizing, $\Sigma_\epsilon\cap B_1(0)$ must be the solution to Plateau's problem. Moreover, it is the unique solution thanks to the fact that the $\Sigma_\epsilon$'s form a foliation (e.g., by the maximum principle or a calibration argument).

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Now, via the above facts, I claim we can construct a small perturbation of the Euclidean metric so that $\Gamma$ bounds a unique area minimizing surface which is smooth. To do so, simply construct a diffeomorphism $\phi:\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ so that $\phi(\Gamma_\epsilon)=\Gamma$. Clearly we can arrange that $\phi$ is close to the identity in whatever sense we want. Now, for the metric $\phi^*\delta$, the solution to the Platau problem for $\Gamma$ is unique and smooth, because it is $\phi(\Sigma_\epsilon)$.