Well, the identity component $H^0$ of $H$ would have to be abelian, since, otherwise, it would have a compact simple component, and the induced map on $\pi_3$ would then be nontrivial.

If $H^0$ has positive dimension and is abelian, then what you are asking is whether $\pi_1(H^0)\to\pi_1(\mathrm{SO}(n))\simeq \mathbb{Z}_2$ is trvial or not.  This does happen, of course.  For example, if $H^0=S^1$ and it represents the zero element in $\pi_1(SO(n))\simeq \mathbb{Z}_2$, then the induced map on homotopy is trivial in $\pi_1$ and hence in all higher cases as well.  This first happens for $n=4$, as there is such an $S^1$ sitting in $\mathrm{SO}(4)$ (in fact, a countably distinct family of them).  It also happens for all $n\ge 4$.