I will prove below that your bound $\frac{\exp\left(C\frac{\log N}{\log \log N}\right)}{A^3}$ (which follows from $\sum_{d\mid N, d > A} \frac{1}{d^3} \le \frac{d(N)}{A^3}$) is optimal at least in the regime $A = N^c$, where $0 < c < 1$ is fixed. (note that we can't have $A$ very small since $\sum_{d\in \mathbb{N}} \frac{1}{d^3} < \infty$).

Put $N = p_1p_2\ldots p_k$ where $k$ is some natural number and $p_j$ are prime numbers. From, say, prime number theorem we have $k = \Theta \left(\frac{\log N}{\log \log N}\right)$. I will construct $\Theta\left(\exp\left(C\frac{\log N}{\log \log N}\right)\right)$ divisors of $N$ in the interval $(A, 2A]$ from what the desired estimate follows. Construction goes as follows:

we choose random subset of primes $p_j$ with $j > [k\left(1 - \frac{1}{100}\min(c^5, (1-c)^5)\right)] = m$ and call their product $d_1$(note that there are already required number of $d_1$'s). It is also easy to see that $d_1 < A$. Then we are doing the following greedy algorithm: initialize $d := d_1$. Lets look at $p_j$ starting with $p_m$ in decreasing order and multiply $d$ by $p_j$ until one more multiplication will make $d$ greater than $A$. Such a moment exists since $p_1p_2\ldots p_m > A$. Call this moment $j$. We have now that $d \le A < p_jd$. If $p_j d \le 2A$ then let $d:=p_j d$ and finish the algorithm. Otherwise by Bertrand's postulate there is some prime $p$ in the interval $(\frac{A}{d}, \frac{2A}{d}]$ and $p < p_j$. Let $d := pd$ and finish the algorithm.

In any case we will find divisor $d\in (A, 2A]$ of $N$ and all of them are obviously different. Thus our claim is proved.