Ryan gives a very nice answer in dimension 3, leaving the higher-dimensional 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 $(n-2)$-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 4-manifolds.

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 higher-dimensional knots, by Bruno Eckmann, begins:

E. Dyer and R. Vasquez proved that the complement of a higher-dimensional knot $S^{n-2} \subset S^n$, $n \ge 4$, is never aspherical unless the knot group is infinite cyclic (and hence, for $n \ge 5$, the imbedding is unknotted). [Footnote referring to papers of Levine, Stallings, Wall, Shaneson.] In the present note we give a simple proof of this fact based on some remarks concerning compact $\partial$-manifolds.

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 cut-and-paste arguments as Ryan describes. In dimension $n=4$, the surgery theory is (a) much more difficult in the continuous category, and (b) non-existent in the smooth/PL category.

The paper The algebraic characterization of the exteriors of certain 2-knots, by Jonathan Hillman, credits Freedman with the result that a 2-knot 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 4-manifold 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 4-manifold 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 4-manifold is completely unknown. So if the smooth Poincare conjecture is sufficiently false, presumably you could have a non-trivial 2-knot in $B$ that becomes trivial in $M$.

Ryan gives a very nice answer in dimension 3, leaving the higher-dimensional 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 $(n-2)$-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 4-manifolds.

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 higher-dimensional knots, by Beno Eckmann, begins:

E. Dyer and R. Vasquez proved that the complement of a higher-dimensional knot $S^{n-2} \subset S^n$, $n \ge 4$, is never aspherical unless the knot group is infinite cyclic (and hence, for $n \ge 5$, the imbedding is unknotted). [Footnote referring to papers of Levine, Stallings, Wall, Shaneson.] In the present note we give a simple proof of this fact based on some remarks concerning compact $\partial$-manifolds.

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 cut-and-paste arguments as Ryan describes. In dimension $n=4$, the surgery theory is (a) much more difficult in the continuous category, and (b) non-existent in the smooth/PL category.

The paper The algebraic characterization of the exteriors of certain 2-knots, by Jonathan Hillman, credits Freedman with the result that a 2-knot 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 4-manifold 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 4-manifold 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 4-manifold is completely unknown. So if the smooth Poincare conjecture is sufficiently false, presumably you could have a non-trivial 2-knot in $B$ that becomes trivial in $M$.