Take a closed $n$manifold $M$ and fix a nice $n$ball $B$ in $M$. How much information about $M$ does the set of knotted $(n2)$spheres of $B$ which are unknotted in $M$ give?
Kevin Walker and a member of the forum (by email) let me know about a flaw in the argument in my previous post. I'll erase it after putting up this post. One of my claims was the if $K$ unknots in $M$ then it unknots in $B$, where $K$ is a knot in a ball $B \subset M$. If $K$ is the unknot in $M$ it bounds a disk $D \subset M$, $K = \partial D$. In dimension $3$ we perturb the disc so that $D \cap \partial B$ is a collection of circles, this allows us to cap off $D \cap B$ with disc(s) parallel to $\partial B$ (an embedded surgery construction), constructing a disc in $B$ that bounds $K$. In higher dimensions this argument runs into trouble as it's not clear if you can perform the required embedded surgeries. I also claimed that the only new isotopy relation in general is that knots in $B$ can be isotopic (in $M$) to their mirror inverse if $M$ is nonorientable. I still suspect this is the case and I think I have an idea for an argument but I should think about it some more before posting. Here is an argument that works in dimension 3: Say $M$ is a compact boundaryless $3$manifold, $K$ and $K'$ contained in a ball $B$ are isotopic in $M$. Perform the connectsum decomposition on $M \setminus K$ and $M \setminus K'$. Since this is a unique decomposition (modulo issues in the nonorientable case), in both cases you get a list of prime factors that are identical except in the decomposition of $M \setminus K$ you get a summand that's $B \setminus K$ capped off with a disc, and in the other case you get a summand $B \setminus K'$ capped off with a disc. By the GordonLuecke theorem $K$ and $K'$ are either isotopic in $B$ or mirror images of each other. I'm still hopeful I have an argument that works in general. I'll edit this post again tomorrow once I'm certain. edit: final exams and deadlines are eating up my time at the moment but I'll get around to another revision... eventually. 


Ryan gives a very nice answer in dimension 3, leaving the higherdimensional case of the question open. I can't discuss the question with as much authority as I would like, but I can start to piece together an answer. My impression is that in any dimension $n$, an $(n2)$knot $K$ that is unknotted in $M$ is already unknotted in $B$ in the continuous category. In the smooth and PL categories, it should be the same except in $n=4$ dimensions, where the question runs into a lack of understanding of smooth (or equivalently PL) structures on 4manifolds. First, my impression from skimming some geometric group theory is that a finitely presented group cannot be an infinite free product. Moreover, that it if it is a finite free product, the factors have unique isomorphism types. If this impression is correct, then you can identify the knot group $\pi_1(S^n \setminus K)$ from the embedded fundamental group $\pi_1(M \setminus K)$. In order to be trivial in $M$, the knot group of $K$ would have to be $\mathbb{Z}$. Then, Aspherical manifolds and higherdimensional knots, by Bruno Eckmann, begins:
My impression is that the surgery methods in dimension $n \ge 5$ used to prove the unknottedness assertion are valid in the smooth, PL, and continuous categories. Dimension $n=3$ is of course a special case where you do not need to consider algebraic topology, but instead prove things with direct cutandpaste arguments as Ryan describes. In dimension $n=4$, the surgery theory is (a) much more difficult in the continuous category, and (b) nonexistent in the smooth/PL category. The paper The algebraic characterization of the exteriors of certain 2knots, by Jonathan Hillman, credits Freedman with the result that a 2knot with knot group $\mathbb{Z}$ is unknotted. If this requires Freedman's work, then I would think that it is open in the smooth/PL case. What would be open is whether the 4manifold with boundary $B^3 \times S^1$ has more than one smooth structure. If it does, I have seen a principle that smooth structures on a 4manifold can merge together when you take a connected sum with manifolds such as $\pm \mathbb{C}P^2$ an $S^2 \times S^2$. I do not know if this principle is established for manifolds with a fundamental group. But at the very least, I have heard that the semigroup of smooth structures on $S^4$ is completely unknown, and its action on the smooth structures on another 4manifold is completely unknown. So if the smooth Poincare conjecture is sufficiently false, presumably you could have a nontrivial 2knot in $B$ that becomes trivial in $M$. 

