Let $0 < \theta < 1$ be a fixed real number, and $0 < c_1 < c_2 < 1$ be real numbers. For a positive integer $n$, let $d_{\theta, c_1, c_2}(n)$ be the number of divisors $k$ of $n$ satisfying
$$\displaystyle c_1 n^\theta < k < c_2 n^\theta.$$
My question is, how large can $d_{\theta, c_1, c_2}(n)$ as a function of $n$? In particular, can $d_{\theta, c_1, c_2}(n)$ be larger than $(\log n)^A$ for any positive number $A$?
For any choice of $C_1,C_2,\theta$ with $\theta \in (0,1)$ and $0 < C_1 < C_2$ (not sure why you said $C_2 < 1$), there are infinitely many $N$ with at least $c\exp\Bigl(c(1-\theta)\frac{\log N}{\log \log N}\Bigr)$ many divisors in $[C_1 N^\theta, C_2 N^\theta]$, where $c$ is an absolute constant.
I slightly adapt this argument.
Let $x$ be large and let $M = \prod_{p \le x} p$. Partition $[1,M]$ into intervals of the form $[K,L]$ with $K \le L < \frac{C_2}{C_1}K$. Since there will be $\approx_{C_2/C_1} \log M$ many intervals, there will be one, call it $[K,L]$, with at least $c\exp(c \frac{\log M}{\log \log M})$ many divisors (for large enough $M$) (any $c < \log 2$ should work).
Now, set $R := \biggl\lfloor \Bigl(\frac{C_2 M^\theta}{L}\Bigr)^{1/(1-\theta)} \biggr\rfloor$, and let $N = M R$. Then, for each divisor $d \in [K,L]$ of $M$, the number $dR$ is a divisor of $N$, and it satisfies $$dR \le LR \le C_2 N^\theta$$ $$dR \ge KR \ge \frac{C_1}{C_2}LR \ge C_1 N^\theta.$$
It remains to note that the number of divisors we found, namely $c\exp(c \frac{\log M}{\log \log M})$, is at most $c\exp\Bigl((1-\theta)c\frac{\log N}{\log\log N}\Bigr)$ since $N \le M^{1/(1-\theta)}$.
