Such an annulus need not exist. For example, consider two circles in $\mathbb{R}^3$ defined by $x^2+y^2 = 1$ and $z = \pm R$. If $R$ is sufficiently large, then there cannot be a minimizing annulus (or, indeed, any minimizing connected surface) with these two circles as boundary.
The reason is the following: First, one can certainly find an annulus with these two surfaces as boundary whose area is as close to $2\pi$ as desired: Just take the two flat disks with these circles as boundary and join them by a very thin tube $x^2+y^2 = \epsilon^2>0$ and $|z|\le R$ for $\epsilon$ very small and smooth the result to get an annulus.
Meanwhile, if there were a connected minimal surface $A$ with these circles as boundary, then $A$ would have to pass through the plane $z=0$ at some point $p = (x_0,y_0,0)$, and hence the ball $B$ of radius $r<R$ centered on $p$ would not meet the two circles. Then the monotonicity formula for minimal surfaces implies that the part of the surface $A$ inside the ball $B$ would have to have area at least $\pi r^2$. Letting $r$ approach $R$, one sees that the area of $A$ would be at least $\pi R^2$.
Thus, if $R^2>2$, then no connected minimal surface with these two circles as boundary can achieve the lower bound of $2\pi$.