demanding $n \geq 1$ cuts out a few counterexamples...
WOW p 7 q 9 r 31 a 0 n 1 1_1: -16 f -12
WOW p 7 q 9 r 31 a 1 n 2 1_1: -16 f -8
WOW p 9 q 11 r 49 a 0 n 1 1_1: -28 f -12
WOW p 9 q 11 r 49 a 1 n 2 1_1: -28 f -12
WOW p 9 q 11 r 49 a 1 n 3 1_1: -28 f -12
WOW p 9 q 11 r 49 a 2 n 5 1_1: -28 f -20
WOW p 9 q 11 r 49 a 2 n 6 1_1: -28 f -24
WOW p 11 q 13 r 71 a 0 n 1 1_1: -44 f -12
WOW p 11 q 13 r 71 a 0 n 2 1_1: -44 f -40
WOW p 11 q 13 r 71 a 1 n 2 1_1: -44 f -20
WOW p 11 q 13 r 71 a 1 n 3 1_1: -44 f -12
WOW p 11 q 13 r 71 a 1 n 4 1_1: -44 f -20
WOW p 11 q 13 r 71 a 2 n 5 1_1: -44 f -36
WOW p 11 q 13 r 71 a 2 n 6 1_1: -44 f -24
WOW p 11 q 13 r 71 a 2 n 7 1_1: -44 f -28
WOW p 11 q 13 r 71 a 3 n 9 1_1: -44 f -36
WOW p 11 q 13 r 71 a 3 n 10 1_1: -44 f -36
WOW p 11 q 15 r 41 a 1 n 2 1_1: -16 f -8
WOW p 13 q 15 r 97 a 0 n 1 1_1: -64 f -12
WOW p 13 q 15 r 97 a 0 n 2 1_1: -64 f -40
WOW p 13 q 15 r 97 a 1 n 2 1_1: -64 f -32
WOW p 13 q 15 r 97 a 1 n 3 1_1: -64 f -16
WOW p 13 q 15 r 97 a 1 n 4 1_1: -64 f -16
WOW p 13 q 15 r 97 a 1 n 5 1_1: -64 f -32
WOW p 13 q 15 r 97 a 2 n 6 1_1: -64 f -40
WOW p 13 q 15 r 97 a 2 n 7 1_1: -64 f -28
WOW p 13 q 15 r 97 a 2 n 8 1_1: -64 f -32
WOW p 13 q 15 r 97 a 2 n 9 1_1: -64 f -52
WOW p 13 q 15 r 97 a 3 n 10 1_1: -64 f -48
WOW p 13 q 15 r 97 a 3 n 11 1_1: -64 f -40
WOW p 13 q 15 r 97 a 3 n 12 1_1: -64 f -48
WOW p 13 q 15 r 97 a 4 n 14 1_1: -64 f -56
jagy@phobeusjunior:~$
(I'm going to use $B^n$ to denote an $n$-dimensional open ball and $D^n$ to denote a closed one.)
Theorem. Let $K$ be a knot. $S^3-\nu(K)\approx S^1\times D^2$ if and only if $K$ is the unknot.
Proof. We will show the stronger statement that $\pi_1(S^3-\nu(K))\approx\mathbb{Z}$ if and only if $K$ is the unknot. Consider a minimal genus Seifert surface $\Sigma$ for $K$, and let $\Sigma'=\Sigma-\nu(K)$. The induced map $\pi_1(\Sigma')\to \pi_1(S^3-\nu(K))$ must be injective, since otherwise by Kneser's lemma there would be a way to compress $\Sigma'$ and reduce its genus. If the knot genus is greater than $0$, then $\pi_1(S^3-\nu(K))$ contains a free group on at least two generators. The rest follows from the fact that the unknot is the unique genus-$0$ knot. $\square$
In fact, Gordon and Luecke proved that if there is an orientation-preserving homeomorphism $S^3-\nu(K)\approx S^3-\nu(K')$, then $K$ and $K'$ are equivalent knots.
Since disks are contractible, the tubular neighborhood $\nu(D)$ of a disk $D$ in $D^4$, regarded as the embedded normal bundle, must be a trivial $\mathbb{R}^2$ bundle over $D$. So, $\nu(D)\approx D\times B^2$. Its closure is $\overline{\nu(D)}\approx D^4$.
The complement $D^4-\nu(D)$ can be rather complicated. While I don't have any examples on hand, it seems that $\pi_1(D^4-\nu(D))\neq \mathbb{Z}$ if $D$ is a ribbon disk for a non-trivial ribbon knot by the van Kampen theorem.
In contrast to the Gordon and Luecke result, I just found a paper by Abe and Tange, "Ribbon disks with the same exterior", where they give a family of inequivalent slice knots such the the complements of a slice disk for each are all diffeomorphic.
Best Answer
You can pick up some of this theory from the likes of Hatcher's notes on 3-manifolds or Hempel's or Jaco's books on 3-manifold topology, or perhaps even Rolfsen's book on knot theory. Lickorish sprinkles some here and there at various points in his book.
The intuition for a ball $B\subset S^3$ meeting $K$ as a trivial ball-arc pair being "essentially unique" is that you can shrink $B$ down so that it can be thought of as a tiny bead on $K$. Then, any two beads on $K$ are equivalent by sliding them around.
A trivial ball-arc pair $(B,\alpha)$ is a "ball $B$ with an unknotted spanning arc $\alpha\subset B$", which more precisely is a pair that is, as a subspace, PL homeomorphic to the pair $$[0,1]\times (D^2,0)=([0,1]\times D^2,[0,1]\times 0).$$ Let $(B,\alpha)\subset (S^3,K)$ be a trivial ball-arc pair.
By Alexander's theorem (that for a PL embedded $2$-sphere $S\subset S^3$, the closure of each component of $S^3-S$ is PL homeomorphic to a $3$-ball with boundary $S$) and Alexander's trick, we can get an isotopy of $S^3$ that isotopes $B$ to a standard $3$-ball in $S^3$ while isotoping $\alpha$ to a diameter of $B$. Alexander's trick uses the PL homeomorphism that characterizes a trivial ball-arc pair.
Once in this position, we can make a knot diagram for $K$ where $\alpha$ is indicated by small arc in the diagram away from any crossings, where $B$ projects onto a small disk with the arc as its diameter. We are allowed to slide this arc around the diagram (crossing crossings) since there is a corresponding isotopy of $S^3$ to implement this move. Reidemeister moves (which correspond to certain isotopies of $S^3$) are allowed at least if the arc is not in the way, but the arc can always be moved to be out of the way. Also, isotopies of the diagram itself correspond to isotopies of $S^3$ as usual. From here, it should be clear that any trivial ball-arc pair is equivalent to any other, since the Reidemeister moves and isotopies of diagrams generate knot equivalence.
I used the Reidemeister moves as a tool to deal with $S^3$ isotopies, but this was not necessary. Other options include (1) taking the ball-arc pair and extending it into a closed tubular neighborhood of $K$, then using the fact that every parameterization of a closed tubular neighborhood is isotopic to any other, and then finishing it off with the fact that any closed interval on $S^1$ is isotopic to any other, (2) using the parameterization of the trivial ball-arc pair to shrink it so that it lies inside a given tubular neighborhood, then seeing the way in which it sits with respect the foliation of the tubular neighborhood by disks to isotope it (depending on the present singularities) into a standard ball with only a maximum and minimum, and any pair of such balls are more-or-less-obviously isotopic. Most of the complexity is keeping track of the arc inside the ball.