The paper below shows that a least area can be found spanning any rectifiable curve. Sometimes the disk is not immersed however, as with a disk bounding a figure 8 in the plane.

J. Hass, Singular curves and the Plateau problem, International J. of Math. 2, (1991) 1-16.

For your example, there is a more straightforward construction of a minimal disk with boundary $\gamma$ that gives a disk that is immersed. You assume that a catenoid spans $C_+$ and $C_-$. The union of this catenoid and the cylinder between $C_+$ and $C_-$ is a torus that bounds a solid torus. The boundary of this piecewise smooth solid torus has mean curvature that points into the solid torus (or at least never out of it) and the angles at the curves where the two surfaces intersect are less than $\pi$. This "mean convex" boundary condition suffices to allow the classic solution of Plateau's problem to be solved for simple curves, as shown in the Meeks Yau papers- see "Topology of Three Dimensional Manifolds and the Embedding Problems in Minimal Surface Theory" William H. Meeks and Shing-Tung Yau, Annals of Mathematics, Vol. 112, No. 3 (Nov., 1980), pp. 441-484.

Now your curve is not embedded on the boundary of the solid torus, but the curve lifts to an embedded curve in a double cover of the solid torus. Solve the Plateau problem there (using the induced metric, also mean convex) and project to get a minimal disk.

For large $R$ this will be the least area disk spanning $\gamma$. It stays close to the cylinder. For small $R$ I think there might be a smaller disk that cuts across the axis.