Yes. I assume that by "compact surface diffeomorphic to a closed annulus" you mean
a bordered surface. Then by definition of a bordered surface, each boundary component has more than one point. Now an open Riemann surface homeomorphic to
an annulus is conformally equivalent to $\{ z: r<|z|<R\}$, where
$0\leq r<R\leq+\infty$. If each boundary component is not a point, we have
the honest annulus with $r=1,R<\infty$.

That a neighborhood of a puncture is not conformally equivalent to
a neighborhood of the non-degenerate boundary component is a consequence of
the elementary theorem on the removable singularity.

EDIT. Let me add some detail. The punctured annuli are characterized by the property that the extremal
length of the family of all homotopically non-trivial curves is zero.
On the other hand, if we have a bordered Riemann surface, homemorphic to a
non-degenerate annulus, we can estimate this extremal length from below.
To do this, we put on our annulus some conformal metric, in which the
length of the boundary components is not zero.
In your case, the conformal structure is defined by a diffeomorphism. Just pull
back the Euclidean metric on $1\leq |z|\leq R$ by your diffeomorphism.

For the definition and properties of
extremal length, see any of the two books of Ahlfors,
Conformal invariants or Lectures on quasiconformal mappings.

An alternative proof can be obtained by referring to Kwak's theorem,
which is the generalization of the removable singularity theorem to maps between
Riemann surfaces. (Ann. Math., 90 (1969) 9-22, or S. Lang, Introduction to complex hyperbolic spaces, Springer, 1987.)