When I read this question, I interpreted "loops" more broadly than seems to have been intended, allowing for the possibility that a loop might cross itself. With this possibility in hand, the argument in Dr Xorile's answer no longer works because
if a loop has N self-intersections then a tiny displacement of that loop must cross the original at least 2N times.
In this broader setting, the answer is
still that the game can never finish
and here is one way to see it:
note first of all that any loop divides the plane into regions that can be coloured with two colours, say red and blue, with no two like-coloured regions adjoining. Now take a look at that last loop (call it L) and consider its relations with earlier loops. As you walk around an earlier loop, every time it crosses from one L-region to another it meets L, so it can only do that twice, so it can only pass through two of the L-regions. And since any two of those earlier loops meet, there must be an L-region they both pass through. I claim that in fact there is a single L-region through which all earlier loops pass. Suppose that's not so, and consider two earlier loops M,N. They meet in, say, region A; perhaps M goes through another region B and perhaps N goes through another region C. Now consider any other loop O. It has to meet either A or B, and it has to meet either A or C. If it doesn't meet A then it must meet both B and C. But this is impossible, because when we do the colouring described earlier B and C have the same colour (whatever colour A isn't), and you therefore can't get from B to C and back to B without crossing L more than twice. So in fact any other loop has to meet region A, and we have found a region through which all loops before L pass. And now we're done, because we can take a loop that goes around the inside boundary of that region very very close to L, which necessarily intersects each earlier loop just twice and L not at all, and then we can perturb it a little to get two intersections with L.