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Edit: This solution is incorrect, but I'm leaving it here because I think it's still interesting.

Here's a solution Scott Morrison and I came up with.

Choose a trivalent graph on the sphere with the vertices connected by segments of great circles such that any slight pertubation of the vertices would result in an a greater total edge length. The really symmetric tetradodecahedron on the sphere is such a graph.^† Now replace each vertex with a "clove hitch vertex" and each edge by a pair of strands:

Introduce some twists along the edges in order to make the whole link into a single knot. Two strands running between a pair of vertices can't be separated without making them longer, so the only way to deform the knot-around-the-sphere is to deform the "underlying graph", and we've chosen the graph so that any deformation would result in strictly larger total edge length.

The end result (without the twists to make a single component) would look like this if you used a tetrahedron instead:

^†If you don't believe that the symmetric tetrahedron is minimal in this way, either prove it in another answer, or ask another question! Edit: As some of you have pointed out, the tetrahedron is not such a minimal graph, and we were being too greedy asking for a minimal graph where the vertices are so far apart. So you have to find some graph which actually is minimal. I think a dodecahedron should do the trick (edit: it doesn't), but I don't know how to prove it.

Edit: This solution is incorrect, but I'm leaving it here because I think it's still interesting.

Here's a solution Scott Morrison and I came up with.

Choose a trivalent graph on the sphere with the vertices connected by segments of great circles such that any slight perturbation of the vertices would result in an a greater total edge length. The really symmetric tetradodecahedron on the sphere is such a graph.^† Now replace each vertex with a "clove hitch vertex" and each edge by a pair of strands:

Introduce some twists along the edges in order to make the whole link into a single knot. Two strands running between a pair of vertices can't be separated without making them longer, so the only way to deform the knot-around-the-sphere is to deform the "underlying graph", and we've chosen the graph so that any deformation would result in strictly larger total edge length.

The end result (without the twists to make a single component) would look like this if you used a tetrahedron instead:

^†If you don't believe that the symmetric tetrahedron is minimal in this way, either prove it in another answer, or ask another question! Edit: As some of you have pointed out, the tetrahedron is not such a minimal graph, and we were being too greedy asking for a minimal graph where the vertices are so far apart. So you have to find some graph which actually is minimal. I think a dodecahedron should do the trick (edit: it doesn't), but I don't know how to prove it.

Edit: This solution is incorrect, but I'm leaving it here because I think it's still interesting.

Here's a solution Scott Morrison and I came up with.

Choose a trivalent graph on the sphere with the vertices connected by segments of great circles such that any slight pertubation of the vertices would result in an a greater total edge length. The really symmetric tetradodecahedron on the sphere is such a graph.^† Now replace each vertex with a "clove hitch vertex" and each edge by a pair of strands:

Introduce some twists along the edges in order to make the whole link into a single knot. Two strands running between a pair of vertices can't be separated without making them longer, so the only way to deform the knot-around-the-sphere is to deform the "underlying graph", and we've chosen the graph so that any deformation would result in strictly larger total edge length.

The end result (without the twists to make a single component) would look like this if you used a tetrahedron instead:

^†If you don't believe that the symmetric tetrahedron is minimal in this way, either prove it in another answer, or ask another question! Edit: As some of you have pointed out, the tetrahedron is not such a minimal graph, and we were being too greedy asking for a minimal graph where the vertices are so far apart. So you have to find some graph which actually is minimal. I think a dodecahedron should do the trick (edit: it doesn't), but I don't know how to prove it.

Edit: This solution is incorrect, but I'm leaving it here because I think it's still interesting.

Here's a solution Scott Morrison and I came up with.

Choose a trivalent graph on the sphere with the vertices connected by segments of great circles such that any slight pertubation of the vertices would result in an a greater total edge length. The really symmetric tetradodecahedron on the sphere is such a graph.^† Now replace each vertex with a "clove hitch vertex" and each edge by a pair of strands:

Introduce some twists along the edges in order to make the whole link into a single knot. Two strands running between a pair of vertices can't be separated without making them longer, so the only way to deform the knot-around-the-sphere is to deform the "underlying graph", and we've chosen the graph so that any deformation would result in strictly larger total edge length.

The end result (without the twists to make a single component) would look like this if you used a tetrahedron instead:

^†If you don't believe that the symmetric tetrahedron is minimal in this way, either prove it in another answer, or ask another question! Edit: As some of you have pointed out, the tetrahedron is not such a minimal graph, and we were being too greedy asking for a minimal graph where the vertices are so far apart. So you have to find some graph which actually is minimal. I think a dodecahedron should do the trick (edit: it doesn't), but I don't know how to prove it.

Edit: This solution is incorrect, but I'm leaving it here because I think it's still interesting.

Here's a solution Scott Morrison and I came up with.

Choose a trivalent graph on the sphere with the vertices connected by segments of great circles such that any slight pertubation of the vertices would result in an a greater total edge length. The really symmetric tetradodecahedron on the sphere is such a graph.^† Now replace each vertex with a "clove hitch vertex" and each edge by a pair of strands:

Introduce some twists along the edges in order to make the whole link into a single knot. Two strands running between a pair of vertices can't be separated without making them longer, so the only way to deform the knot-around-the-sphere is to deform the "underlying graph", and we've chosen the graph so that any deformation would result in strictly larger total edge length.

The end result (without the twists to make a single component) would look like this if you used a tetrahedron instead:

^†If you don't believe that the symmetric tetrahedron is minimal in this way, either prove it in another answer, or ask another question! Edit: As some of you have pointed out, the tetrahedron is not such a minimal graph, and we were being too greedy asking for a minimal graph where the vertices are so far apart. So you have to find some graph which actually is minimal. I think a dodecahedron should do the trick (edit: it doesn't), but I don't know how to prove it.

Edit: This solution is incorrect, but I'm leaving it here because I think it's still interesting.

Here's a solution Scott Morrison and I came up with.

Choose a trivalent graph on the sphere with the vertices connected by segments of great circles such that any slight pertubation of the vertices would result in an a greater total edge length. The really symmetric tetradodecahedron on the sphere is such a graph.^† Now replace each vertex with a "clove hitch vertex" and each edge by a pair of strands:

Introduce some twists along the edges in order to make the whole link into a single knot. Two strands running between a pair of vertices can't be separated without making them longer, so the only way to deform the knot-around-the-sphere is to deform the "underlying graph", and we've chosen the graph so that any deformation would result in strictly larger total edge length.

The end result (without the twists to make a single component) would look like this if you used a tetrahedron instead:

^†If you don't believe that the symmetric tetrahedron is minimal in this way, either prove it in another answer, or ask another question! Edit: As some of you have pointed out, the tetrahedron is not such a minimal graph, and we were being too greedy asking for a minimal graph where the vertices are so far apart. So you have to find some graph which actually is minimal. I think a dodecahedron should do the trick (edit: it doesn't), but I don't know how to prove it.