Extending $S^1$ to an embedding of $D^2$ in $\mathbb{R}^3$ Consider the specific embedding $f : S^1 \to \mathbb{R}^3$ given by, say, the unit circle in the $xy$-plane. Suppose further that this embedding is contained within a $3$-dimensional, simply-connected embedded submanifold of $\mathbb{R}^3$, say $M$. Then can this specific embedding be extended to an embedding $\hat{f} : D^2 \to M$? From simple-connectedness of $M$, it can be extended to a continuous function $D^2 \to M$.
If $f$ was some arbitrary embedding then the answer is no, because $f(S^1)$ could be knotted in $\mathbb{R}^3$. But I specifically chose an $f$ that is not knotted in $\mathbb{R}^3$.
 A: Just to clarify the setup: you have a smooth embedding $f:S^1\to \mathbb{R}^3$ such that $f(S^1)$ bounds an embedded disk (this is the definition of unknotted, and if $f(S^1)$ lies in the $xy$ plane then the 2D Schoenflies theorem gives such a disk), and you have a simply connected $3$-dimensional submanifold $M\subset \mathbb{R}^3$ with $f(S^1)\subset M$.
Let's assume that $M$ is a closed subset of $\mathbb{R}^3$, meaning it is a submanifold with boundary.  This is satisfied if $M$ is the closure of the result of removing codimension-$0$ submanifolds from $\mathbb{R}^3$.
Let's add one additional assumption that each boundary component of $M$ is compact.  ($M$ could be something like the the regular neighborhood of an embedded $4$-regular tree -- the Cayley graph of the rank-two free group -- and the boundary of this is an infinitely punctured $S^2$.)  The Whitehead manifold suggests that there might be difficulties with the non-compact case, and the edit history of this answer contains an incorrect argument for the non-compact case.
Here is a characterization of these manifolds:
Claim. Every compact boundary component $\Sigma\subset \partial M$ is diffeomorphic to $S^2$.
Proof. First, note that $\Sigma$ is orientable since it has an induced orientation from $M$.  If $\Sigma$ is not $S^2$, then the induced map $\pi_1(\Sigma)\to \pi_1(M)$ has nontrivial kernel since $M$ is simply connected.  By the loop theorem there is a properly embedded disk $(D^2,S^1)\to (M,\Sigma)$ (meaning the disk meets the boundary transversely) -- see Prof. Calegari's notes.  Let $M'\subset M$ be the result of cutting out a tubular neighborhood of this disk (i.e., compressing the boundary along the disk).  If $M'$ is disconnected, then $\pi_1(M)$ is a free product of the fundamental groups of each component hence each component is simply connected, then by induction on genus of $\Sigma$ both sides after compression are $S^2$'s, so re-joined $\Sigma$ is an $S^2$. If $M'$ were connected, then notice that $M$ is homotopy equivalent to $M'\vee S^1$, which would mean $\pi_1(M)\neq 0$, a contradiction. QED
By Alexander's theorem (a.k.a. the 3D Schoenflies theorem), each of these boundary components bounds an embedded ball in $\mathbb{R}^3$.  One can then deduce that  $M$ is $B^3$ minus balls or $\mathbb{R}^3$ minus balls, but we won't use this.  I just figure it's good to know what we are dealing with.  In fact, the claim applies to any compact boundary component of any oriented simply connected $3$-manifold.
Now extend the original loop to a map $f:D^2\to \mathbb{R}^3$.  While still $f(S^1)\subset M$, it might or might not be the case that $f(D^2)\subset M$.  By a small isotopy, we may assume $f$ meets $\partial M$ transversely.
Consider $f(D^2)\cap \partial M$, which is a collection of simple closed curves, from the perspective of $\partial M$.  Take an innermost curve $\gamma$, meaning there is a disk $C\subset\partial M$ such that $f(D^2)\cap C=\gamma$.  Replace the definition of $f$ on the interior of the curve $f^{-1}(\gamma)$ by $C$, and isotope $f$ away from $C$ into the interior of $M$.  This removes at least one curve of intersection in $f(D^2)\cap C$.  We may inductively continue to remove innermost curves until none are left, thus leaving us with an embedding $f:D^2\to M$ with $f(S^1)$ the original curve, as required.
(Note that you can use a variant of "half-lives half-dies" to argue that genus $g>0$ boundaries have nonzero symplectic forms from the inclusion on first homology.)
