By Venkov's proof of the Evens-Venkov theorem, if you take a faithful representation $G\rightarrow SO(m)$, then the cohomology $H^*(BG)$ is a finitely generated module for the image of $H^*(BSO(m))$. But the situation is quite different if you only allow yourself Euler classes, and allowing Euler classes of all representations doesn't help much.
There are insufficiently many such classes in general to come close to generating the whole cohomology ring.
Consider the group $G$ defined as the affine group $G=AGL(1,q)$ for $q$ a power of a prime $p$, say $q=p^n$. This is the group of all self-maps of the field $F_q$ of the form $x\mapsto ax+b$ with $a\neq 0$. This $G$ has a normal subgroup isomorphic to $(C_p)^n$ and quotient a cyclic group of order $p^n-1$. The irreducible complex representations of this group are easily computed: there are $p^n$ 1-dimensional representations and one other one of dimension $p^n-1$. The single large representation is already realized over the rationals. So the irreducible real representations are one of dimension $p^n-1$ and lots of others of dimension at most 2. The low-dimensional representations factor through the projection from $G$ onto $C_{q-1}$. So by taking Euler classes of irreducible real representations, you get a subring of $H^*(BG)$ in which only one of your generators has order divisible by $p$. The other generators have image 0 under the restriction map $H^*(BG)\rightarrow H^*(B(C_p)^n)$.
By the Evens-Venkov theorem, the cohomology of $(C_p)^n$ is a finitely-generated module for $H^*(BG)$ under this map. When $n>1$, $H^*(B(C_p)^n)$ cannot be a finitely-generated module for a subring generated by 1 element, so in this sense Euler classes do not come close to generating $H^*(BG)$.
One more remark: the smallest group to which my argument applies is the affine group $AGL(1,4)$, which is isomorphic to the alternating group $A_4$ mentioned in some of the other solutions.