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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.

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 $f(\partial X)$.

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 $f(\partial X)$.

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.

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.

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 $f(\partial X)$.

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 $f(\partial X)$.