Here we will provide a necessary and sufficient condition for $I_k$ and $J_k$ to be bounded away from $0$.

We have
\begin{equation}
I_k=E\|a_k^{1/2}X\|^n\,1(a_k^{1/2}\|X\|<r_k),
\end{equation}
where $X$ is a centered Gaussian random vector in $\mathcal H$ with covariance operator $C$, and
$n$ is fixed (as clarified in the OP's comment).

We will have to distinguish two cases:

Case 1, when $r_k/a_k^{1/2}$ is bounded away from $0$;

Case 2, when $r_k/a_k^{1/2}\to0$ (as $k\to\infty$).

In Case 1, we have $r_k/a_k^{1/2}\ge c$ for some $c>0$ and all $k$, so that
$$I_k=a_k^{n/2}E\|X\|^n\,1(\|X\|<r_k/a_k^{1/2}) \\
\ge a_k^{n/2}E\|X\|^n\,1(\|X\|<c).$$
On the other hand, clearly $I_k\le a_k^{n/2}E\|X\|^n$.

Thus, in Case 1, $I_k$ is bounded away from $0$ iff $a_k$ is bounded away from $0$.

Consider now Case 2, which is more interesting. Here the key is

**Lemma 1:** $P(\|X\|^2<2x)\gtrsim(1+b)P(\|X\|^2<x)$ (as $x\downarrow0$) for some $b>0$, depending on the distribution of $\|X\|$, where $A\gtrsim B$ means $A\ge(1+o(1))B$.

This lemma will be proved at the end of this answer.

At this point, let us let $x_k:=r_k^2/a_k$, so that $0<x_k\to0$ and use Lemma 1 to write
\begin{equation}
\begin{aligned}
I_k&=a_k^{n/2}E\|X\|^n\,1(\|X\|^2<x_k) \\
&\ge a_k^{n/2}E\|X\|^n\,1(x_k/2\le\|X\|^2<x_k) \\
&\ge a_k^{n/2}(x_k/2)^{n/2}\,P(x_k/2\le\|X\|^2<x_k) \\
&\gtrsim a_k^{n/2}(x_k/2)^{n/2}\,b\,P(\|X\|^2<x_k) \\
&=Cr_k^n\,P(\|X\|^2<r_k^2/a_k),
\end{aligned}
\end{equation}
where $C:=(1/2)^{n/2}\,b$.
On the other hand, clearly $I_k\le a_k^{n/2}x_k^{n/2}\,P(\|X\|^2<x_k)
=r_k^n\,P(\|X\|^2<r_k^2/a_k)$.

Thus, in Case 2, $I_k$ is bounded away from $0$ iff $r_k^n\,P(\|X\|^2<r_k^2/a_k)$ is bounded away from $0$.
An exact asymptotic for the so-called small-ball probability $P(\|X\|^2<x)$ (with $x\downarrow0$) is known -- see e.g. Theorem 2, but the asymptotic expression depends on $x$ and the eigenvalues of the covariance operator $C$ of $X$ in a complicated manner.

The consideration of $J_k$ is similar to that of $I_k$, but much simpler; essentially, dealing with $J_k$, we may assume that we are dealing again with $I_k$ but only for $\mathcal H=\mathbb R$.

It remains to provide

*Proof of Lemma 1:* Note that $\|X\|^2$ equals $\sum_{j\ge1}c_j Z_j^2$ in distribution, where the $c_j$'s are the positive eigenvalues of the covariance operator $C$ of $X$ and the $Z_j$'s are independent standard normal random variables (r.v.'s), so that $\sum_{j\ge1}c_j=E\|X\|^2<\infty$. So, for real $x>0$,
\begin{equation}
P(\|X\|^2<x)=\int_0^x G(x-y)\,dH(y),
\end{equation}
where $G$ and $H$ are the c.d.f.'s of $c_1 Z_1^2$ and $\sum_{j\ge2}c_j Z_j^2$, respectively.
Note that the r.v. $c_1 Z_1^2$ has a gamma distribution with shape parameter $1/2$, so that $G(2u)\gtrsim(1+b)G(u)$ for some $b>0$ and $u\downarrow0$. So, for $x\downarrow0$,
\begin{equation}
P(\|X\|^2<2x)=\int_0^{2x} G(2x-y)\,dH(y)
\ge\int_0^x G(2x-2y)\,dH(y) \\
\gtrsim \int_0^x (1+b)G(x-y)\,dH(y)=(1+b)P(\|X\|^2<x).\quad\Box
\end{equation}