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Edit : here is the beginning of an explanation of "why" $m$ is square in this example ($393129=627^2$). Recall that an integer $n \geq 1$ is a square if and only if its number of divisors is odd (consider the involution $d \mapsto \frac{n}{d}$ on the set of divisors of $n$). If $n$ is odd, then all its divisors are odd, so that $n$ is a square if and only if $\sigma(n)$ is odd. Now consider $N=(1+q+pq)m$ as above. The condition on $\sigma(N)/N$ implies that $\sigma(N)$ is even but not divisible by $4$.

If we assume that $1+q+pq$ and $m$ are coprime, then $\sigma(N)=\sigma(1+q+pq) \sigma(m)$, so the reasoning above shows that $1+q+pq$ or $m$ is a square (but not both). If $1+q+pq=\alpha^2$ then $\alpha \equiv \pm 1 \pmod{q}$ so that $\alpha \geq q-1$, which leads to a contradiction. Thus $m$ is a square (it is possible to show further that $1+q+pq$ is a prime times a square).

If $1+q+pq$ and $m$ are not coprime, the situation is more intricate (this is what happens in the example I found : we had $\operatorname{gcd}(1+q+pq,m)=9$). Let $m'$ be the largest divisor of $m$ which is relatively prime to $1+q+pq$. Put $m=\lambda m'$. Then $\lambda(1+q+pq)$ or $m$ is a square. I don't see an argument for excluding the first possibility, but at least if $\lambda$ is a square then so is $m$.

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I did a little computer search and I think I found an example of an odd immaculate group.

I searched for groups of the form $G=(C_q \rtimes C_p) \times C_N$ with odd primes $p,q$ such that $p | q-1$ and $N$ an odd integer satisfying $(N,pq)=1$. Using Tom's notations and results, we have

\begin{equation*} \frac{D(G)}{|G|} = \frac{D(C_q \rtimes C_p)}{|C_q \rtimes C_p|} \cdot \frac{D(C_N)}{|C_N|} = \frac{1+q+pq}{pq} \cdot \frac{\sigma(N)}{N} \end{equation*} where $\sigma(N)$ denotes the sum of divisors of $N$. We want $\frac{\sigma(N)}{N} = \frac{2pq}{1+q+pq}$. Since the last fraction is irreducible, $N$ has to be of the form $N=(1+q+pq)m$ with $m$ odd. I found the following solution :

\begin{equation*} p=7, \quad q=127, \quad m=393129. \end{equation*} This gives the immaculate group $G=(C_{127} \rtimes C_7) \times C_{399812193}$, which has order $|G| = 355433039577 = 3^4 \cdot 7 \cdot 11^2 \cdot 19^2 \cdot 113 \cdot 127$.