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 $(q,k)\equiv (1,1)\pmod 3$.
The inequality
$$\frac{2p^k + D(p^k)}{p^k + D(p^k)} \leq \frac{2m^2}{m^2 + D(m^2)}$$
holds if and only if $$p\ge 13$$
Proof :
We see that
$$\frac{2p^k + D(p^k)}{p^k + D(p^k)} \leq \frac{2m^2}{m^2 + D(m^2)}\tag1$$
is equivalent to
$$ (p-8)p^{2k+1}+8p^{2k}+2p^k(3p-4)+1\ge 0\tag2$$
because
$$\begin{align}(1)&\iff 2m^2(p^k+D(p^k))-(2p^k+D(p^k))(m^2+D(m^2))\ge 0
\\\\&\iff 2m^2p^k+2m^2(2p^k-\sigma(p^k))-2p^km^2-2p^k(2m^2-\sigma(m^2))
\\&\qquad \qquad -m^2(2p^k-\sigma(p^k))-(2p^k-\sigma(p^k))(2m^2-\sigma(m^2))\ge 0
\\\\&\iff 4p^k\sigma(m^2)+m^2\sigma(p^k)-8p^km^2\ge 0
\\\\&\iff 4p^k\frac{2p^km^2}{\sigma(p^k)}+m^2\sigma(p^k)-8p^km^2\ge 0
\\\\&\iff 8p^{2k}+\sigma(p^k)^2-8p^k\sigma(p^k)\ge 0
\\\\&\iff 8p^{2k}+\bigg(\frac{p^{k+1}-1}{p-1}\bigg)^2-8p^k\frac{p^{k+1}-1}{p-1}\ge 0
\\\\&\iff 8p^{2k}(p-1)^2+(p^{k+1}-1)^2-8p^k(p^{k+1}-1)(p-1)\ge 0
\\\\&\iff (p-8)p^{2k+1}+8p^{2k}+2p^k(3p-4)+1\ge 0\end{align}$$
Now,
If $p\ge 13$, then we see that $(2)$ holds.
If $p\lt 13$, i.e. $p=5$, then $(2)$ is equivalent to $5^k(7\cdot 5^{k}-22)\le 1$ which is impossible.
In conclusion, we have
$$\frac{2p^k + D(p^k)}{p^k + D(p^k)} \leq \frac{2m^2}{m^2 + D(m^2)}\iff p\ge 13$$
Best Answer
I've got some partial results.
This answer proves the following claims :
Claim 1 : If $N=2^k$, then $\frac{N}{D(N)}$ is an almost perfect number.
Claim 2 : If $N = 2^k(2^{k+1} + 2^t - 1)$ where $2^{k+1} + 2^t - 1$ is an odd prime and $0\lt t\le k$, then $\frac{N}{D(N)}$ is not an almost perfect number.
Claim 3 : If $N$ is not of the form $2^k$ and $N\le 49024$, then $\frac{N}{D(N)}$ is not an almost perfect number.
Claim 4 : If $\frac{N}{D(N)}=2^k$ where $k\ge 1$ with $D(N)\not=1$, then
$D(N)$ is even.
$D(N)$ has at least three distinct prime factors.
$D(N)$ is not of the form $2pq$ where $p,q$ are distinct odd primes.
$D(N)$ is not of the form $2^2pq$ where $p,q$ are distinct odd primes such that $(p,q)\not\equiv (1,1)\pmod 4$.
$D(N)\ge 90$
Claim 1 : If $N=2^k$, then $\frac{N}{D(N)}$ is an almost perfect number.
Proof : We have $D(N)=2^{k+1}-(2^{k+1}-1)=1$ and $$\sigma\bigg(\frac{N}{D(N)}\bigg)-\bigg(\frac{2N}{D(N)}-1\bigg)=\sigma(2^k)-\sigma(2^k)=0\qquad\blacksquare$$
Claim 2 : If $N = 2^k(2^{k+1} + 2^t - 1)$ where $2^{k+1} + 2^t - 1$ is an odd prime and $0\lt t\le k$, then $\frac{N}{D(N)}$ is not an almost perfect number.
Proof : If $N = 2^k(2^{k+1} + 2^t - 1)$ where $2^{k+1} + 2^t - 1$ is an odd prime, then we get $$D(N)=2^{k+1}(2^{k+1} + 2^t - 1)-(2^{k+1}-1)(2^{k+1} + 2^t)=2^t$$ from which we have $$\begin{align}&\sigma\bigg(\frac{N}{D(N)}\bigg)-\bigg(\frac{2N}{D(N)}-1\bigg) \\\\&=\sigma\bigg(2^{k-t}(2^{k+1} + 2^t - 1)\bigg)-(2^{k+1}-1)(2^{k-t+1}+1) \\\\&=(2^{k-t+1}-1)(2^{k+1}+2^t)-(2^{k+1}-1)(2^{k-t+1}+1) \\\\&=(2^{k-t+1}+1)(1-2^t) \\\\&\not=0\qquad\blacksquare\end{align}$$
Claim 3 : If $N$ is not of the form $2^k$ and $N\le 49024$, then $\frac{N}{D(N)}$ is not an almost perfect number.
Proof : In the following, red numbers are of the form $2^k$, and blue numbers are of the form $2^k(2^{k+1} + 2^t - 1)$ where $2^{k+1} + 2^t - 1$ is an odd prime and $t\le k$ where $a_n$ is the $n$-th deficient-perfect number and $$f(N)=\sigma\bigg(\frac{N}{D(N)}\bigg)-\bigg(\frac{2N}{D(N)}-1\bigg)$$Note here that $$\text{$\dfrac{N}{D(N)}$ is an almost perfect number}\iff f(N)=0$$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\ \hline a_n&\color{red}1&\color{red}2&\color{red}4&\color{red}8&\color{blue}{10}&\color{red}{16}&\color{red}{32}&\color{blue}{44}&\color{red}{64}&\color{red}{128}&\color{blue}{136}&\color{blue}{152}&\color{blue}{184}&\color{red}{256}&\color{red}{512}\\ \hline f(a_n)&0&0&0&0&\not=0&0&0&\not=0&0&0&\not=0&\not=0&\not=0&0&0 \\\hline \end{array}$$ $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n&16&17&18&19&20&21&22&23&24&25&26\\ \hline a_n&\color{blue}{752}&884&\color{red}{1024}&\color{red}{2048}&\color{blue}{2144}&\color{blue}{2272}&\color{blue}{2528}&\color{red}{4096}&\color{red}{8192}&\color{blue}{8384}&\color{blue}{12224} \\ \hline f(a_n)&\not=0&-189&0&0&\not=0&\not=0&\not=0&0&0&\not=0&\not=0 \\\hline \end{array}$$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n&27&28&29&30&31&32&33&34&35\\ \hline a_n&\color{red}{16384}&17176&18632&18904&\color{red}{32768}&\color{blue}{32896}&\color{blue}{33664}&\color{blue}{34688}&\color{blue}{49024} \\ \hline f(a_n)&0&-111&-1863&-2205&0&\not=0&\not=0&\not=0&\not=0 \\\hline \end{array}$$
Claim 4 : If $\frac{N}{D(N)}=2^k$ where $k\ge 1$ with $D(N)\not=1$, then
$D(N)$ is even.
$D(N)$ has at least three distinct prime factors.
$D(N)$ is not of the form $2pq$ where $p,q$ are distinct odd primes.
$D(N)$ is not of the form $2^2pq$ where $p,q$ are distinct odd primes such that $(p,q)\not\equiv (1,1)\pmod 4$.
$D(N)\ge 90$
Proof : Let $D(N)=m$. Then, we have $$\begin{cases}2N-\sigma(N)=m \\\\\frac{N}{2N-\sigma(N)}=2^k\end{cases}\implies\sigma(2^km)=m(2^{k+1}-1)\tag1$$ for some $k\ge 1$.
Suppose that $m\ge 3$ is odd. Then, we have $$(1)\implies\sigma(2^k)\sigma(m)=m(2^{k+1}-1)\implies \sigma(m)=m$$ which is impossible.
Suppose that $m=2^s$ where $s\ge 1$. Then, we have $$(1)\implies 2^{k+s+1}-1=2^s(2^{k+1}-1)$$which is impossible since LHS is odd while RHS isn't.
Suppose that $m=2^st^u$ where $s\ge 1$ and $t$ is an odd prime with $u\ge 1$. Then, we have $$(1)\implies (2^{k+s+1}-1)(1+t+\cdots +t^u)=2^st^u(2^{k+1}-1)$$ There exists a positive integer $v$ such that $$1+t+\cdots +t^u=2^sv\tag2$$ to have $$(2^{k+s+1}-1)v=t^u(2^{k+1}-1)$$ There eixsts a positive integer $w$ such that $$2^{k+1}-1=vw\tag3$$ to have $$2^{k+s+1}=t^uw+1\tag4$$ Multiplying the both sides of $(3)$ by $2^s$ and using $(2)(4)$ give $$1-2^s=(1+t+\cdots +t^{u-1})w$$ which is impossible since LHS is negative while RHS isn't.
So, we see that $D(N)$ has at least three distinct prime factors.
Suppose that $m=2pq$ where $p,q$ are distinct odd primes. Then, we get $$(1)\implies (2^{k+1}-1)(p+1)(q+1)=2pq(2^{k+1}-1)$$ which is impossible since LHS is divisible by $4$ while RHS isn't.
Suppose that $m=2^2pq$ where $p,q$ are distinct odd primes and $(p,q)\not\equiv (1,1)\pmod 4$. Then, we get $$(1)\implies (2^{k+3}-1)\cdot\frac{p+1}{2}\cdot\frac{q+1}{2}=pq(2^{k+1}-1)$$ which is impossible since LHS is even while RHS isn't.
Hence, it follows that $D(N)\ge 2\cdot 3^2\cdot 5=90$. $\quad\blacksquare$