Here is a general way to construct a large family of bendings of the cylinder $M$, with nonplanar boundaries, via Alexandrov's isometric embedding theorem. All these examples will be convex.

First note that the geodesic curvature of the boundary components of $M$ is zero. Let $D_1$ and $D_2$ be any pairs of convex planar disks with smooth positively curved boundaries $\partial D_1$, $\partial D_2$ which have the same length as the boundary components of $M$. Gluing $D_i$ along the boundaries of $M$ yields a a closed surface $\overline M$ with a metric that has everywhere nonnegative curvature, because geodesic curvatures of $\partial D_i$ are nonnegative. Thus, by Alexandrov's isometric embedding theorem, $\overline M$ admits an isometric embedding into $R^3$ as a convex surface $\overline M'$. Let $M'$ and $D_i'$ be the images of $M$ and $D_i$ in $\overline M'$.

I claim that $M'$ cannot have planar boundary components when $D_1$ and $D_2$ are non-congruent. Indeed, the only time when $M'$ has planar boundaries is when $D_1$ and $D_2$ are congruent, and the corresponding points on their boundaries are glued to the end points of the same line of ruling of $M$.
To see this note that if $D_1'$ is planar then the principal normals of $\partial D_1'$ (which are non-vanishing by the positive curvature assumption on $\partial D_i$) must lie in the plane of $D_1'$. But the principal normals of $\partial D_1'$ must be orthogonal to $M'$ since boundary components of $M'$ are geodesics (being a geodesic is an isometric invariant). So $M'$ has to be orthogonal to the plane of $D_1'$.

Thus, if $D_1'$ is planar, then all tangent planes of $M'$ along $\partial D_1'$ are orthogonal to the plane of $D_1'$. But since $M'$ is convex, its tangent planes are support planes. So $M'$ has to lie inside a right cylinder, say $S_1$ generated by $D_1'$. Similarly, if $D_2'$ is planar, then $M'$ has to lie inside a right cylinder $S_2$ generated by $D_2'$. It follows then that, if $D_1'$ and $D_2'$ are both planar, then $S_1=S_2\supset M'$. In particular $D_1'$ and $D_2'$ are congruent, lie in parallel planes, and are directly "above" each other (i.e., the orthogonal projection of $D_1'$ into the plane of $D_2'$ coincides with $D_2'$).

**Note:** It is important in the above construction that $\partial D_i$ be smooth. Indeed when $D_i$ are triangles, then $M'$ will form the sides of an antiprism, with bases $D_i'$.