Consider the metric on $\mathbb R^2$ that is rotationally symmetric metric outside a compact set, namely, it is $dr^2+m(r)^2 d\phi^2$ for $r>R>0$. Here $m$ is a positive function on $[R,\infty)$.
The area form at points with $r>R$ is $dA=m(r)drd\phi$, so the surface has finite area if and only if $m$ is integrable on $[R,\infty)$. The total curvature of the rotationally symmetric end is $$\int_R^\infty -\frac{m^{\prime\prime}}{\!\!\! m} dA=-2\pi\int_R^\infty m^{\prime\prime} dr=2\pi\left(m'(R)-\lim_{r\to\infty} m'(r)\right).$$ It is easy to find examples where the limit on the right hand side does not exist but $m$ is integrable. If the limit exists and $m$ is integrable, the limit is zero, and the total curvature of the end is $2\pi m^\prime(R)$.
The metric is arbitrary if $r<R$, and the total curvature of this region can be computed with the usual Gauss-Bonnet for surfaces with boundary. The geodesic curvature of the boundary is easy to compute from the $r\ge R$ side (I don't remember the answer).
EDIT: As Willie Wong points out by Gauss-Bonnet every smooth metric on the region $\{r<R\}$ will have the same total curvature. So just extend $m$ to a smooth function on $[0,R]$ so that $m(r)=r$ near $0$ and consider the metric $dr^2+m(r)^2d\phi^2$ for all $r>0$. Its metric completion is smooth at the origin. (More generally, the metric is smooth at the origin if and only if $m^\prime(0)=1$ and $m$ extends to an odd smooth function on $\mathbb R$). Now the above computation gives total curvature as $2\pi(m^\prime(0)-m^\prime(\infty))=2\pi(1-m^\prime(\infty))$ and if $m^\prime(\infty)$ exists and the area is finite, the total curvature is $2\pi$.