A stable domain $D$ in a minimal surface $S\subset \mathbb{R}^3$ is a domain for which the area-functional $A(t):=\int_{S_t}dS_t$ has non-negative second derivative, i.e. $A''(0)\geq 0$, for all compactly supported variatons on $D$. We consider normal variations of the form $F:S\times ]-\epsilon,\epsilon[\rightarrow \mathbb{R}^3: (p,t)\mapsto p+tf(p)N_p$, with $f:S\rightarrow \mathbb{R}$ a compactly supported function on $S$ and $N_p$ a local Gauss map on $S$. We denote $S_t$ the varied surface at time $t$, (so $S_0=S$).
It can be shown that a minimal surface is a critical point of the area functional, i.e. $A'(0)=0$, so therefore a stable domain in a minimal surface reaches a minimum of the area functional.
My question: Can we then say that a stable domain $D$ in a minimal surface $S$ is area minimizing among all surfaces with the same boundary $\partial D$ and contained in the cylinder $\partial D \times \mathbb{R}$? Or when and under which conditions is a stable domain in a minimal surface area minimizing among all surfaces with the same boundary $\partial D$?
The short answer to your question is no, a stable domain $D$ does not in general minimise area among all surfaces with the same boundary.
Let me elaborate: for simplicity assume that $D$ is orientable, so that the Gauss map is defined on the whole of $D$. Let $D_t$ be the surfaces resulting from a variation along $f \in C_c^1(D)$ and denote $A(D_t)$ their respective areas. At least for small $t$, the surfaces $D_t$ are going to be embedded in $\mathbb{R}^3$ also. The following definitions are standard: $D$ is called stable if $\frac{\mathrm{d^2}}{\mathrm{d}t^2} A(D_t) \geq 0$ for all variations $f \in C_c^1(D)$, and is called area-minimizing if it has the least area of all surfaces with the same boundary. (Sometimes this is also called absolutely area-minimizing to distinguish with the case where one restricts oneself to surfaces with the same genus as $D$, but this distinction is not important here.)
As $S$ is minimal, we have $\frac{\mathrm{d}}{\mathrm{d}t} A(D_t) = 0$. However, this does not show that the function $t \mapsto A(D_t)$ reaches a minimum at $t = 0$, even when $\frac{\mathrm{d^2}}{\mathrm{d}t^2} A(D_t) \geq 0$. The problem is roughly the same as for functions: the higher-order variations of $A(D_t)$ may dominate. Nobody is to say that $A(D_t)$ does not behave like $t^3$ near $0$, for instance.
Of course, the surfaces $D_t$ all have the same boundary as $D$, so this answers your second question in the negative: not only is a stable domain $D$ in general not area-minimizing among surfaces with the same boundary, but it might even be that $A(D_t) < A(D)$.
The question also asks whether stable domains are area-minimizing "in the cylinder $\partial D \times \mathbb{R}$". Though I understand the idea behind this, I do not really see a way of making sense of this. How would you embed this cylinder in $\mathbb{R}^3$? If for instance you chose to embed it using the normal to $S$ at every point in $\partial D$, you would generally get self-intersections. In any case, any reasonable definition of surfaces "in the cylinder $\partial D \times \mathbb{R}$" should include the surfaces $D_t$ constructed above, so the answer would still be no.
You also ask about conditions under which a stable minimal surface would automatically be area-minimizing. The questions seems to me a little bit like asking when a function $f: (-\epsilon,\epsilon) \to \mathbb{R}$ with $f'(0) = 0$ and $f''(0) \geq 0$ has a minimum at $0$. In other words, I do not know any non-trivial conditions on $D$ that would ensure this, but I would be happy to think about it some more if you could elaborate on what you were conjecturing.