Does the Image of a Loop under Contraction contain a Disc? Suppose we have a point $x \in X$, where $X$ is a Hausdorff space.  Let $J$ be an embedded copy of the circle $\mathbb{S}^1 \subset X$ containing $x$, and let $F$ be a deformation retract of $J$ onto $x$ in $X$.  Define the "trace" $tr(F)$ of $F$ to be all points $y \in X$ such that there are $j \in J, t \in [0,1]$ satisfying $F(j,t) = y$.  In other words, it's the "full image of $J$ under $F$" in $X$.
Let $D$ denote the closed unit disc in the plane, and $S$ its boundary.

Is there an embedding $i: D \rightarrow tr(F)$ such that $x \in i(S)$?

The question of whether there's a continuous extension to $D$ has been answered a couple times on here, but I'm wondering if contracted loops carve out a disc.  Note that this makes no sense for general contractible spaces, e.g. trees.  This is also weaker than asking if every such loop bounds a disc, which should be impossible by gluing a couple wild arcs in $\mathbb{R}^3$ together (someone correct me if I'm wrong on that, not my area; they might have to be extra-wild ones).
If it's false, what are some sufficient conditions?  Maybe metric $X$ and strong deformation retract?

This is not an answer, but too long for a comment. You ask whether
for each contraction $F : J \times I \to X$ of $J$ to some $x \in J$ there exists an embedded disk $D' \subset tr(F)$ such that $x \in \partial D'$.
This may depend not only on $J$, but also on the choice of $F$. As an example consider any knot $J \subset \mathbb R^3$ ( as wild as you want). Take a disk $D' \subset \mathbb R^3$ with $x \in \partial D'$. There clearly is a homotopy $H: J \times I \to \mathbb R^3$ rel. $x$ such that $H_0 = id$ and $H_1(J) = \partial D'$. Now combine it with the standard contraction of $D'$. Then $F$ has the desired property.
This leaves of course open whether a given $F$ has the same property. However, you should consider what you really want to know: Whether each $F$ is "nice" or whether there exists a "nice" $F$.
