On OP's request, I am converting my comment into an answer.
Also, I'm going to add some more thoughts at the end of this answer.
I noticed the following :
(1) In the case of even perfect numbers, we have
$$\frac 74\color{red}{\le} I(\sigma(2^p−1))=2−\frac{1}{2^p}\lt 2$$
from which we see that $\sigma(2^p−1)$ is deficient.
(2) If $\sigma(q^k)=u^sv^t$ where $u\lt v$ are primes such that $5\color{red}{\le} v$, then $I(\sigma(q^k))<\dfrac 32\cdot \dfrac{v}{v−1}\lt 2$, so $\sigma(q^k)$ is deficient.
(3) If $\sigma(q^k)=u^s\cdot 3^t$ where $u$ is an integer (not necessarily a prime) such that $\gcd(u,3)=1$, then $\dfrac u2\ (=m)$ is odd with $s=1$, and $$I(\sigma(q^k))=\dfrac{3\sigma(m)(3^{t+1}-1)}{2m\cdot 3^t\cdot 2}=\underbrace{\dfrac 34\left(3−\dfrac{1}{3^t}\right)}_{\ge 2}\cdot \underbrace{\dfrac{\sigma(m)}{m}}_{\ge 1}\ge 2$$ so $\sigma(q^k)$ is not deficient.
In the following, I'm going to add some more thoughts.
(4) One can prove that if $(q,k)$ satisfies either $q\equiv 2\pmod 3$ or $(q,k)\equiv (1,2)\pmod 3$, then $\sigma(q^k)$ is not deficient.
Proof :
If $q\equiv 2\pmod 3$, then we have
$$\sigma(q^k)=1+q+\cdots +q^k\equiv (1-1)+(1-1)+\cdots +(1-1)\equiv 0\pmod 3$$since $k$ is odd.
Also, if $(q,k)\equiv (1,2)\pmod 3$, then we have
$$\sigma(q^k)=1+q+\cdots +q^k\equiv 1+1+\cdots +1\equiv k+1\equiv 0\pmod 3$$
So, in either case, we get $\sigma(q^k)\equiv 0\pmod 3$.
Since we have $\sigma(q^k)\equiv 2\pmod 4$, there are positive integers $s,t$ such that $$\sigma(q^k)=2s\cdot 3^t$$where $s$ is odd satisfying $\gcd(s,3)=1$. Then, we have
$$I(\sigma(q^k))=\frac{3\sigma(s)(3^{t+1}-1)}{2s\cdot 3^t\cdot 2}=\underbrace{\dfrac 34\left(3−\dfrac{1}{3^t}\right)}_{\ge 2}\cdot \underbrace{\dfrac{\sigma(s)}{s}}_{\ge 1}\ge 2$$
from which we see that $\sigma(q^k)$ is not deficient.
So, the remaining cases are $(q,k)$ satisfying either $(q,k)\equiv (1,0)\pmod 3$ or $(q,k)\equiv (1,1)\pmod 3$.
Too long to comment :
Let $c:=\dfrac{\ln(4/3)}{\ln(13/9)}$. Using $(I(m^2))^{c} < I(m)$, one can get
$$m\le \dfrac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{c}}\implies D(m)\lt D(p^k)\tag1$$
which is better than $$m < p^k \implies D(m) < D(p^k).$$
Proof :
$(I(m^2))^c\lt I(m)$ is equivalent to
$$\bigg(\frac{\sigma(m^2)}{m^2}\bigg)^c\lt\frac{\sigma(m)}{m}\iff \sigma(m)\gt m\bigg(\frac{\sigma(m^2)}{m^2}\bigg)^c=m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$$
Using $\sigma(m)\gt m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$, we get
$$D(p^k)-D(m)=D(p^k)-2m+\sigma(m)\gt D(p^k)-2m+m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$$
So, we can say that if
$$D(p^k)-2m+m\bigg(\dfrac{2}{I(p^k)}\bigg)^c\ge 0\tag2$$
then $D(m)\lt D(p^k)$ where note that
$$(2)\iff m\le \frac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^c}$$ So, we can say that $(1)$ holds.
To see that $(1)$ is better than $m < p^k \implies D(m) < D(p^k)$, it is sufficient to prove that
$$p^k\lt \frac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{c}}\tag3$$
We have
$$\begin{align}(3)&\iff p^k\lt \frac{p^{k+1}-2p^k+1}{(p-1)\bigg(2-\bigg(\dfrac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\bigg)}
\\\\&\iff p^k(p-1)\bigg(2-\bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\bigg)\lt p^{k+1}-2p^k+1
\\\\&\iff 2-\bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\lt \frac{p^{k+1}-2p^k+1}{p^k(p-1)}
\\\\&\iff \bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\gt 2-\frac{p^{k+1}-2p^k+1}{p^k(p-1)}
\\\\&\iff \bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\gt \frac{p^{k+1}-1}{p^k(p-1)}
\\\\&\iff \frac{2p^k(p-1)}{p^{k+1}-1}\gt \bigg(\frac{p^{k+1}-1}{p^k(p-1)}\bigg)^{1/c}
\\\\&\iff 2\gt \bigg(\frac{p^{k+1}-1}{p^{k}(p-1)}\bigg)^{(c+1)/c}
\\\\&\iff 2^{c/(c+1)}\gt \frac{p^{k+1}-1}{p^{k}(p-1)}
\\\\&\iff 2^{c/(c+1)}p^k(p-1)-p^{k+1}+1\gt 0
\\\\&\iff p^k\underbrace{\bigg((2^{c/(c+1)}-1)p-2^{c/(c+1)}\bigg)}_{\text{positive}}+1\gt 0\end{align}$$
which does hold. So, $(3)$ holds.
Best Answer
Too long to comment :
One can prove that $$D(m^2)\lt D(p^k)\iff m^2\lt \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)}{2(p-1)(p^k-1)}$$
Proof :
We have $$\begin{align}&D(p^k)-D(m^2) \\\\&=2p^k-\sigma(p^k)-2m^2+\sigma(m^2) \\\\&=2p^k-\sigma(p^k)-2m^2+\frac{2p^km^2}{\sigma(p^k)} \\\\&=2p^k-\frac{p^{k+1}-1}{p-1}-2m^2+\frac{2p^km^2(p-1)}{p^{k+1}-1} \\\\&=\frac{(p-1)(p^{k+1}-1)(2p^k-2m^2)-(p^{k+1}-1)^2+2p^km^2(p-1)^2}{(p-1)(p^{k+1}-1)} \\\\&=\frac{ p^{2 k + 2}- 2 p^{2 k + 1} + 2 p^k - 1-2(p^{k + 1} - p^k - p +1)m^2}{(p-1)(p^{k+1}-1)} \\\\&=\frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)-2(p-1)(p^k-1)m^2}{(p-1)(p^{k+1}-1)}\end{align}$$ from which we get $$\begin{align}D(m^2)\lt D(p^k)&\iff \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)-2(p-1)(p^k-1)m^2}{(p-1)(p^{k+1}-1)}\gt 0 \\\\&\iff (p^{k+1}-1)(p^{k+1}-2p^k+1)\gt 2(p-1)(p^k-1)m^2 \\\\&\iff m^2\lt \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)}{2(p-1)(p^k-1)}\end{align}$$