Suppose you had such a decomposition. Pick a circle $C_1$. By Schoenflies it bounds a disc $D_1$. Inductively, pick a circle $C_n$ contained in $D_{n-1}$ such that the area (measure, if you like) of the disc $D_n$ it bounds is at most half the area of the area of $D_{n-1}$.
This is the key step in the proof, so we should carefully see why we can do this. For notational convenience I'm going to call $D=D_{n-1}$.
First note that any disc properly contained in the interior of $D$ has strictly smaller area (its complement in the interior of $D$ is open hence has positive area). Now suppose I could not find a disc of arbitrarily small area bounded by one of our circles; then let the infimum of the areas of discs bounded by our circles be $t>0$. As above we see that we cannot actually achieve $t$, or else we would be able to achieve areas smaller than it by looking at circles inside the disc of area $t$. We will contradict this by showing that we can represent $t$.
Pick a sequence $S_n$ of circles with area at most $t+1/n$. Passing to a suvsequence if necessary and invoking the finite area of the disc I can assume the $S_n$ are nested downwards. Take the intersection of the discs they bound. By Cantor's intersection theorem this is nonempty; pick a point $x$ in it; because the circle containing $x$ must have been in the disc $S_n$ bounds for all $n$, that circle is contained in the infinite intersection; and so is the disc that circle bounds. But this disc must have area at most $t+1/n$ for all $n$, hence have area at most $t$, as desired.
The above discussion proved, recall, that we may always pick a circle $C_n$ contained in the previous, such that the area of the disc it bounds is at most half the previous. Now let's apply the previous argument again: take the intersection of all the $D_n$, it contains a disc, that disc has positive area, which is nonsense since its area is at most $\text{Area}(D_1)/2^n$ for all $n$. This contradicts our most crucial assumption: the existence of a decomposition into circles! Thus proven what we wanted to prove.
I see no way to do this using cohomological arguments. If you demand that the circles are a foliation of the plane it's much easier to prove impossibility.