The first answer to your question is no. There are many acyclic spaces with nonvanishing $\pi_1$. Since generalized homology is a stable invariant, and the suspension of an acyclic space is contractible (exercise), this also means that any generalized cohomology theory vanishes on such a space. For an explicit example, take the classifying space of a perfect group.

But now you'll say it's a $\pi_1$ issue, and that's sort of true. If $E_n(X) =0$ for *every* generalized homology theory, then that means that $E_*(X) = 0$ for any generalized homology theory, including ordinary homology. Indeed: if $E_*(-)$ is a homology theory, so is $E_{*+n}$. But if you're willing to stick to, say, connective cohomology theories (i.e. $E_k(pt) = 0$ for $k<0$), then there are simply connected counterexamples. Indeed: $\pi_3S^2 = \mathbb{Z}$ but $H_3(S^2) = 0$. So let $X = S^2_{\mathbb{Q}}$ denote the rationalization of the 2-sphere, then for any homology theory $E$ with $E_{-1} \otimes \mathbb{Q} = 0$ we get $E_3(X) = 0$ but $\pi_3X = \mathbb{Q}$. Indeed, $\Sigma^{\infty}_+X$ is just $\Sigma^2H\mathbb{Q}$ so $E_3(X) = \pi_{-1}(H\mathbb{Q} \wedge E) = 0$.