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).