You and I decide to play a game:

To start off with, I provide you with a frictionless, perfectly spherical sphere, along with a frictionless, unstretchable, infinitely thin magical rope. This rope has the magical property that if you ever touch its ends to each other, they will stick together and never come apart for all eternity. You only get one such rope, but you are allowed to specify its length.

Next, I close my eyes and plug my ears as you do something to the rope and the sphere. When you are done with whatever you have decided to do, you give me back the sphere and rope. Then I try my best to remove the rope from the sphere (i.e., make the smallest distance from a point on the rope to a point on the sphere at least 1 meter). Of course, since the rope is not stretchable, the total length of the rope cannot increase while I am trying to remove it from the sphere.

If I succeed in removing the rope from the sphere, I win. Otherwise, you win. Who has the winning strategy?

EDIT: To clarify, Zeb is looking for an answer with a finite length, piecewise smooth rope, and the sphere should be rigid.

Without loss of generality we can assume that rope is everywhere tangent to the sphere.
Then some infinitesimal Möbius tranform of its surface will shorten the wrapping length while preserving the crossing pattern (and therefore will loosen the rope without causing it to pass through itself).
Once it is proved, moving in this direction will eventually allow the sphere to escape.
Proof. let $u$ be conformal factor.
Since Möbius tranform preservs total area $\oint u^2=1$ .
Thus, $\oint u<1$.
It follows that for a sutable rotation of $S^2$, we get length decreasing family of Möbius tranforms.
Comments

The same proof works for link made out of 3 circles.

It is easy capture sphere in a link from 4 circles. (4 strings go around 4 faces of regular tetrahedron on $S^2$ and link at the vertexes as in the answer of Anton Geraschenko, the first picture)

BTW, Can one capture a convex body in a knot?


