Is it possible to have a properly embedded annulus $A$ in a genus two handlebody $V$ such that the two boundary curves of $A$ in $\partial V$ represent different isotopy classes of curves on $\partial V$? (I should also mention that in the case I care about, the curves bounding $A$ cannot be meridians, i.e. cannot bound embedded disks in $V$). My intuition says that this is not possible, but maybe I'm just not visualizing things correctly. And I haven't yet thought up a good way to prove this either way.

If the answer is yes, such an annulus exists, can we say anything about whether the boundary curves of $A$ in $\partial V$ are separating/non-separating? For instance, could you have both boundaries be non-separating (as is possible in genus 3)? Or could you have one boundary separating and one non-separating?

I should mention that I mean the following by "properly embedded": $X$ is properly embedded in $Y$ via $f:X \to Y$ if $f(\partial X) = f(X) \cap \partial Y$, and $f(X)$ is transverse to $\partial Y$ at any point of $f(\partial X)$.

Edit: the two answers below are great. But I realized as I was working out these example in my specific case that there is one important detail that I didn't realize was so important: the separating curve must also be disk-busting (i.e. there is no meridian disjoint from the curve). Unfortunately, this means that I cannot apply Kevin's construction because there would be no disjoint meridian; and for Sam's construction, I think the curve has to be parallel to the boundary (?), or at least a curve parallel to the boundary is an example of a disk-busting curve.