For this answer I will consider knots to be links (with one component).
In general the answer is "no". For example, consider $K = 4_1$, the figure-eight knot. Let $X = S^3 - K$. Neither the double nor triple cover of $X$ is a link complement. One of the four-fold covers of X is a link complement, but the other is not. One of its five-fold covers is a link complement, but the other three are not. None of its six-fold covers are link complements. [All of these computations are done in SnapPy - we either find a homeomorphism to a link complement or we find torsion in $H_1$.]
As we take larger and larger covers, we generically expect torsion to appear in $H_1$; such covers do not embed in $S^3$. On the other hand, the figure-eight does have infinitely many covers which are link complements. This is because the "good" four-fold cover is the complement of the link $10^2_{138}$ which has an unknotted component.
I believe that the story will be much the same for any non-trivial knot.