If a manifold $M$ is a double branched cover it admits a non-trivial involution and thus its mapping class group contains an element of order $2$. But in general, one expects a knot complement (or any other manifold) to not have any symmetries. In practice, this can for example be checked using SnapPy.

On the other hand, if one has a manifold $M$ that has an order $2$ element $f$ in its complement (as for example some torus knot complements), then one can take the quotient $M/f$. If $M/f$ is $D^3$ then $M$ is a double branched cover over $D^3$. If $M/f$ is not $D^3$ for all involutions $f$ then $M$ is not a double branched cover over $D^3$ (but over a different manifold).

If a manifold $M$ is a double branched cover it admits a non-trivial involution and thus its mapping class group contains an element of order $2$. But in general, one expects a knot complement (or any other manifold) to not have any symmetries. In practice, this can for example be checked using SnapPy.

On the other hand, if one has a manifold $M$ that has an order $2$ element $f$ in its mapping class group (as for example some torus knot complements), then one can take the quotient $M/f$. If $M/f$ is $D^3$ then $M$ is a double branched cover over $D^3$. If $M/f$ is not $D^3$ for all involutions $f$ then $M$ is not a double branched cover over $D^3$ (but over a different manifold).