After some thinking I understood that existence result for minimal surface bounded by an arbitrary Jordan curve is sufficient for my goals (so I do not need arbitrary jet to prove what I want.) Thanks!

Thanks a lot, Igor. I did know the result (at least locally) since the proof of Douglas for the existence of the minimal surface, or at least the version of the proof I know survives if the target space is not flat; but the precise reference is very important

I do not understand the question, Joseph (and would love if you give more details) but let me still try. I understood your question such that your ''going to be a counterexample'' curve stays from from one side of some closed geodesic and is closer and closer to it for big times. Then, one may think that your ''going to be a counterexample'' lies on a variation of your closed geodesic, i.e., is controlled by Jacobi vector field. By Jacobi vector field is controlled by the equation involving curvature and for positive curvatur it vanishes in finite time so something goes wrong

I possibly misunderstood your question since the trivial answer is to compute the length of all such geodesics, compare, and choose the shortest one, or in fact your programm should already do it since the length is simply the time $s$ one need to go along the geodesic starting at the first point to hit the second point divided by the length of the initial velocity vector.

Is you connection the Levi-Civita connection of a Riemannian metric? Otherwise your way is essentially the only one