EDIT - I've rewritten my previous answer in an attempt to remove everything except the answer to your question. All submanifolds are assumed to be smooth.

Suppose that $M$ is a closed, connected, oriented, irreducible three-manifold (and $M$ is not the three-sphere). Suppose that $K$ is a knot in $M$. Then $\pi_2(M - K)$ is non-trivial if and only if $K$ is contained in an embedded three-ball $B^3 \subset M$.

The proof is an exercise using Alexander's theorem (every embedded two-sphere in $S^3$ bounds balls on both sides) and the sphere theorem. Hempel's book and Hatcher's notes on three-manifolds are standard references for the necessary background material.

EDIT - I've rewritten my previous answer in an attempt to remove everything except the answer to your question. All submanifolds are assumed to be smooth.

Suppose that $M$ is a closed, connected, oriented, irreducible three-manifold (and $M$ is not the three-sphere). Suppose that $K$ is a knot in $M$. Then $\pi_2(M - K)$ is non-trivial if and only if $K$ is contained in an embedded three-ball $B^3 \subset M$.

The proof is an exercise using Alexander's theorem (every embedded two-sphere in $S^3$ bounds balls on both sides), the sphere theorem, and the Poincaré conjecture. Hempel's book and Hatcher's notes on three-manifolds are standard references for the necessary background material.

It is not true in general that $M - K$ has trivial $\pi_2$. Consider the case where $K \subset B^3 \subset M$ where $B^3$ is an embedded three-ball. Then $\partial B$ will be trivial if and only if $M$ is the three-sphere. However, this is the only way to create $\pi_2$, if $M$ is closed, connected, oriented, and irreducible.

You should read about sphere theorem. Hempel's book and Hatcher's notes on three-manifolds are standard references, as well.

EDIT - I've rewritten my previous answer in an attempt to remove everything except the answer to your question. All submanifolds are assumed to be smooth.

Suppose that $M$ is a closed, connected, oriented, irreducible three-manifold (and $M$ is not the three-sphere). Suppose that $K$ is a knot in $M$. Then $\pi_2(M - K)$ is non-trivial if and only if $K$ is contained in an embedded three-ball $B^3 \subset M$.

The proof is an exercise using Alexander's theorem (every embedded two-sphere in $S^3$ bounds balls on both sides) and the sphere theorem. Hempel's book and Hatcher's notes on three-manifolds are standard references for the necessary background material.