Here's the argument I know that avoids spectral sequences, based on the
little-known space $G/N(T)$.
In between $T$ and $G$ is $N(T)$. Note that $EG$ "is an" $ET$ and $EN(T)$,
since it's contractible and $T,N(T)$ act freely on it, so we can
identify $BT, BN(T)$ with $EG/T, EG/N(T)$.
Now consider the two maps $EG/T \to EG/N(T) \to EG/G$, with fibers
$W = N(T)/T$ and $G/N(T)$ respectively. The first case divides by a
free action of $W$, so we can identify
$H^\ast(BN(T);{\mathbb Q}) = H^\ast(BT; {\mathbb Q})^W$ by pushing and pulling.
(Actually we only need to invert $|W|$, and generally less; for $G=U(n)$
it's true over $\mathbb Z$.) In particular, there is only
even cohomology.
So let's look at the space $G/N(T) = (G/T)/W$. The space $G/T$ has a Bruhat
decomposition, hence only even-degree cohomology (even over $\mathbb Z$),
which you can prove via Morse theory on a generic adjoint orbit if you don't
want to bring in algebraic geometry, and its Euler characteristic is $|W|$.
Hence the space $(G/T)/W$ has (rationally) only even-degree cohomology,
and Euler characteristic $1$. So it has the rational cohomology of a point!
For $G=SU(2)$ this space is ${\mathbb RP}^2$.
By a particularly trivial application of Leray-Hirsch (which I think
is the only remainder of the spectral sequence argument Mark Grant gave),
$H^\ast(EG/G; {\mathbb Q}) \cong H^\ast(EG/N(T); {\mathbb Q})$.
(Oops: I guess this answer isn't so different from Ralph's.)
Best Answer
For the group $SU(2)=S^3$ we just have $H^*(BSU(2);\mathbb{Z})=\mathbb{Z}[c_2]$ (where $c_2\in H^4$). More generally, for all $n$ we have \begin{align*} H^*(BU(n);\mathbb{Z}) &= \mathbb{Z}[c_1,\dotsc,c_n] \\\\ H^*(BSU(n);\mathbb{Z}) &= \mathbb{Z}[c_2,\dotsc,c_n] \\\\ H^*(BSp(n);\mathbb{Z}) &= \mathbb{Z}[p_1,\dotsc,p_n] \end{align*} with $c_i\in H^{2i}$ and $p_i\in H^{4i}$.
Now let $V$ be the tautological $3$-plane bundle over the space $X=BSO(3)$. This has Stiefel-Whitney classes $w_2\in H^2(X;\mathbb{Z}/2)$ and $w_3\in H^3(X;\mathbb{Z}/2)$. There is also a Bockstein element $v=\beta(w_2)\in H^3(X;\mathbb{Z})$ (which satisfies $2v=0$) and a Chern class $c=c_2(\mathbb{C}\otimes V)\in H^4(X;\mathbb{Z})$. The mod two reduction map $\rho$ satisfies $\rho(v)=Sq^1(w_2)=w_3$ and $\rho(c)=w_2^2$. If I've got everything straight, one can check using the Bockstein spectral sequence that $$ H^*(BSO(3);\mathbb{Z}) = \mathbb{Z}[v,c]/(2v). $$
It is not possible to be similarly explicit about $H^*(BSO(n);\mathbb{Z})$ for general $n$ (although $H^*(BSO(n);\mathbb{Z}/2)$ and $H^*(BSO(n);\mathbb{Q})$ are fairly straightforward).