Let $M$ be a complete Riemannian 3-manifold and $\gamma \subset M$ a simple closed curve that bounds a *least-area disc* $D$ - a disc that minimizes the area among all discs bounded by $\gamma$.
Let $Conv(\gamma)$ be the convex hull of $\gamma$, i.e, the smallest convex subset of $M$ containing $\gamma$. When $M = \mathbb R^3$ (with the standard Euclidean metric), it is a well-known result that $D \subseteq Conv(\gamma)$.

*Question 1: What (topologcial, metric) conditions need to be satisfied on $M$, along with possible restrictions to $\gamma$, so that $D \subseteq Conv(\gamma)$ holds true ?*

I only know of partial results to this question, such as when $M$ is equipped with a complete metric of bounded sectional curvature and positive injectivity radius, then the above will hold true for all sufficiently small simple closed curves $\gamma$. This is a consequence of a result proven by Joel Hass and Peter Scott in this paper (Theorem 6.3). I am very interested in whether this has been generalized even further as of today.

Note also that when $M = \mathbb R^3$ and $X \subset M$ is any subset, $Conv(X)$ is simply the set of all convex combinations of points in $X$. Orangeskid, a user over at stackexchange has brought to my attention in his answer to a similiar question I posed a week ago that this, along with the convexity of the norm function, is enough to show that \begin{equation} \text{Diam}(Conv(X)) = \text{Diam}(X) \end{equation} always holds true. This means that on $\mathbb R^3$, taking the convex hull of a set does not increase the diameter.

*Question 2: What (topological, metric) conditions need to be satisfied on $M$, so that $\text{Diam}(Conv(X)) = \text{Diam}(X)$ holds true for sufficiently nice subsets $X \subset M$ ?*

The problem here lies in the fact that a general $3$-manifolds lacks a vector space structure, thus both the feasible description of a convex hull, as well as the "convexity" of the metric are not available any more. Any help is appreciated.