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 $n\ge 4$.
As far as classification goes: Given $n$, there will be a countable number of codimension $1$ subtori $T^{r-1}$ in a maximal torus $T^r\subset\mathrm{SO}(n)$ (where $r$, the rank is the greatest integer in $n/2$) such that the inclusion $T^{r-1}\hookrightarrow T^r$ induces an inclusion $\pi_1(T^{r-1})\subset \pi_1(T^r)$ so that all the elements of $\pi_1(T^{r-1})$ are zero in $\pi_1(\mathrm{SO}(n))$. Then the abelian groups that you want are the ones that are conjugate to a subgroup of such a sub-maximal torus.