This is not a complete answer to the question, and I don't know if a complete answer is written down anywhere in the literature. In the first revison of the answer I tried to adress all the 10 comments that my previous answer recieved. The second revison contains a conjecture (that I am 99% sure of) describing the complete answer to this question.

The first point is that the classification of symplectic surfaces can not be simpler than the classification of surfaces up to a diffeo. And the classification up to a diffeo of non-compact surfaces is quite a delicate subject. It is desribed for example in the paper http://www.jstor.org/pss/1993768, 1963, ON THE CLASSIFICATION OF NONCOMPACT SURFACES, IAN RICHARDS. In particular one phenomena apears here -- a certain ideal boundry of the surface. This idel boundary is a totally disconnected, compact separable space. I guess a good illustration will be a disk from which we throw away a Cantor set on the x axis, Cantor set been the ideal boundary.

Let us give now some examples that illustrate the additional phenomena that happen for non-compact surfaces, if we take in account the symplectic form. 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, and 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.

CONJECTURE. Here is the conjecture telling what should be the complete answer to the question.

Take a non-compact surface. Then the set of symplectic structures of infinite area on it is in one to one correspondence with closed non-empty subsets of its ideal boundary.
For every bounded A>0 there is a unique symplectic strucutre on the surface with given area.

In the case of a surface with puncutres, the ideal boundary is just the union of punctures. Below the statement of this conjecture is proven for some simplest examples of surfaces. I think, that the general case should not be much different.

For the simple examples that I have descirbed the proof should be ALMOST identical to Moser's argument. Indeed the simplest examples have the following property:
these surfaces can be decomposed in a countable union of compact pieces, all of which apart from one piece are annuli, and one piece is a compact surface with a boundary. Now, on a compact surface with a boundary (as well as on the cylinder) the symplectic form is uniquely defined by its area -- this can be done by Moser argument (we need that in a neighbrohood of each boundary the symplectic form is strandard, which is automatic in our case). Now the symplectomorphism can be constructed inductively.

Consider for example the case of $R^2$ of infinite area. We can take any exostion of $R^2$ by cylinders. For example we indroduce some coordinates on $R^2$, and conisder cylinders of radiuce $n,n+1$. We don't care what is the exact expression of w. The crucial thing for us is that the sum of areas tends to infinity. Now we take the standard $R^2$ and take a decompositon in concetnric cylinders of needed area. Then define the symplectomorphism from the standrd $R^2$ cylinder by cylinder. For the class of surfaces, that I dealed with same thing should work. Though I am not sure that this is the best "proof" of the statement.

I think it is not hard to prove this conjecture as well in the case of the complement to a Cantor set in the unit disk.