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:
I think $k =1, \alpha \in (0,1)$ should suffice but I didn't think through each step very carefully to verify this.)
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.)
For your question in the comment, I think the following is true:
Theorem: Fix $R_0>0$ and suppose that $||g-g_E||_{C^{k,\alpha}} \leq \epsilon=\epsilon(R_0)$. Then there's a unique $g$-minimizer for any circle of radius $R\leq R_0$.
Proof: If not there's $g\to g_E$ in $C^{k,\alpha}$ and circles $C$ with radius $R\leq R_0$ where things fail. You can translate $C$ to the origin and rescale so that they have radius $1$. The rescaled metrics $g'$ still converge in $C^{k,\alpha}$ (this is where we used $R\leq R_0$). Thus, the previous proof applies.
