Let $K$ be a knot. For an embedding of $K$ into $\mathbb{R}^3$, let $N_1$ be the tubular (open) neighborhood of $K$ in $\mathbb{R}^3$. Similarly, one can embed $K$ in $S^3$ with corresponding tubular neighborhood $N_2$.
Is it true that the spaces $\mathbb{R}^3\setminus N_1$ and $S^3\setminus N_2$ are homotopy equivalent?
No, clearly not. Take the unknot for example. Then $S^3 \setminus N_2$ is homotopy equivalent to a circle $S^1$, whereas $\mathbb{R}^3 \setminus N_1$ is homotopy equivalent to a wedge sum of a circle with a sphere $S^1 \vee S^2$.
In fact $\mathbb{R}^3 \setminus N_1$ is equal to $S^3 \setminus N_2$ with a point removed. Removing a point from a compact 3-manifold with boundary always changes its homotopy type ($M \setminus \{\mathrm{pt}\} \simeq M \vee S^2$).