Below I sketch the proof of the following theorem:

Theorem: Suppose that $\Sigma^n\subset (M^{n+1},g_0)$ is smooth and uniquely area-minimizing relative to it's boundary $\Gamma : = \partial\Sigma$ and is strictly stable (no nontrivial Jacobi fields with Dirichlet boundary conditions). Then if $g$ is a sufficiently small $C^{k,\alpha}$ perturbation of $g_0$, there's a unique area minimizer with boundary $\Gamma$ for the metric $g$ (and it's still smooth and strictly stable).

Remarks:

1. I think $k =1, \alpha \in (0,1)$ should suffice but I didn't think through each step very carefully to verify this.)

2. Note that this theorem actually proves the Euclidean one since if $\gamma,\gamma'$ are nearby boundaries, then you can find $\Phi : \mathbb{R}^n\to \mathbb{R}^n$ diffeomorphism (small) taking $\gamma'$ to $\gamma$. Then the $\gamma'$ solution (for $g_E$) is a $\gamma$ solution for $\Phi^*g_E$.

Proof: Suppose that $g\to g_0$ and $\Sigma_1,\Sigma_2$ are distinct $g$-minimizers with boundary $\Gamma$ (singular is fine). Since $\Sigma$ is smooth then Allard's theorem (interior and boundary) imply that $\Sigma_1,\Sigma_2$ converge smoothly to $\Sigma$ all the way up to the boundary. In other words, there's $u_1,u_2$ on $\Sigma$ so that $\textrm{graph}(u_i) = \Sigma_i$ and $u_i |_{\partial\Sigma} = 0$ (where the graph is taken using the normal exponential map of $g_0$) and $u_i\to 0$ in some $C^{k',\alpha}$ space (which you have to work out depending on $k$). Write $H(u,g)$ for the mean curvature of the graph (as above) of $u$ with respect to $g$. Note that $H(u_i,g) = 0$ by assumption. Thus, $$ 0 = H(u_2,g)-H(u_1,g) = D_1H(u_1,g) (u_2-u_1) + O(||u_2-u_1||_{C^2}^2) $$ (we used Taylor's theorem). The error is uniform with respect to the metric.

Now, divide by $||u_2-u_1||_{C^2}$ to get that $w:=(u_2-u_1)/||u_2-u_1||_{C^2}$ satisfies $$ 0 = D_1H(u_1,g) w + o(1) $$ Using Schauder estimates we can get that $||w||_{C^{2,\alpha}} = O(1)$ (you have to check that the $o(1)$ term is bounded in $C^\alpha$ and that $D_1(u_1,g)w$ is a uniformly elliptic operator. Note that as $g\to g_0$ and $u_1\to 0$, $D_1H(u_1,g)$ limits to the Jacobi operator on $\Sigma$ with respect to $g_0$ (this is the definition of the Jacobi operator $J=D_1H(0,g_0)$. Moreover, by the $C^{2,\alpha}$ bounds for $w$, we get that the limit $w_0$ is $C^2$ and nonzero and solves $Jw_0 = 0$ (with Dirichlet boundary conditions).

This is a contradiction, proving uniqueness for $g$ minimizers, $g$ close to $g_0$. Strict stability follows from the fact that the first Dirichlet eigenvalue of $J$ is continuous as the metric and minimizer varies.

For your second question about the result being uniform for all disks, this cannot be true. Here is a strange proof (probably there is a simpler way):

Construct a non-flat $g$ on $\mathbb{R}^3$ rotationally symmetric with scalar curvature $> 0$ and so that $g=g_E + O(r^{-1})$ at infinity (along with derivatives). This is called an asymptotically flat metric. It turns out that you can do this with $||g-g_E||_{C^k}$ as small as you like. (You can also do this while arranging that there are no closed minimal surfaces in $(\mathbb{R}^3,g)$). This can all be accomplished by looking at the scalar curvature of $dr^2 + \varphi(r)^2 g_{S^2}$ and playing around with the resulting differential inequality.)

Thus, it suffices to prove that for this metric $g$ fixed, for $R$ sufficiently large, the circle of radius $R$ in the $xy$-plane has non-unique minimizers. Suppose the minimizer was unique. Then, by symmetry it must be the disk in the $xy$-plane. (The symmetry of the manifold already shows this is a minimal surface, but it might not be the minimizer; if not the minimizer then $z\mapsto -z$ takes the minimizer to a second one!).

If each disk of radius $R$ was a minimizer then the $xy$-plane will be a minimizer (on compact sets). However, the proof of the positive mass theorem shows that the scalar curvature would vanish along the plane. (Alternatively, see here and the references contained within.)