Let me try to give you some examples that illustrate the phenomena that happen for non-compact surfaces. First of all there is the simplest case when the surface has finite topological type. In this case we have two topological invariants, the fundamental group, a free group on $n$ generators plus the number of punctures (or boundary components) $m$. In this case a complete classification of symplectic forms can be given. Either the surface has a bouned are $A$, in this case this area is the only invariant. Or it has an infinite area. In this case there are $m$ types of surfaces. Namely for every boundary (or puncture) we can check if it has on open neighborhood that his finite area, or not. The number of components near which the area is unbounded can be any between $1$ and $m$.
Example. Consider $S^2$ with two delited points. Then either it has finite area, or it is symplectomorphic to an infinite cylinder $S^1\times R^1$ with the form $ds \wedge dt$, or it is symplectomorphic to $R^2\setminus 0$, with the form $dx\wedge dy$.
If the number of punctures is countable, then already the topological classification becomes complicated. For example, you can throw from the unit disk in $R^2$ the sequence of points with coordinates $(\frac{1}{n},0)$. But if every puncture has a neighborhood that is diffeomorphic to a punctured disk, then the situation should be very similar to what I have described above. Namely 1) the area can be bounded. 2) The number of UNBOUNDED punctures for wich every neighborhood has infinite area is bounded, in such surfaces are enumerated by natural numbers. 3) The number of unbounded punctures is infinite, in this case we just need to cound the number of bounded punctures, that can be finite of inifinite.
Proof should be identical to Moser's argument.
Finally the number of punctures can be uncountable. For example you can take a disk an throw away the Cantor set. I am not avear if in this case one can give a reasonable topological classification of such surfaces.