I'll attempt to sketch a proof that this is true. First, it is convenient to apply the map $z\mapsto \log z$, which maps annular regions in question to thin $2\pi$ - periodic vertical strips $S_r$ and $\mathbb{S}_{r'}$ respectively. I denote the mapping between them by $\varphi_r=u+iv$. The circle $\partial \mathbb{D}$ is mapped to the vertical line $l=\{\Im m\, z=0\}$. Note that $u$ is the unique bounded harmonic function in $S_r$ with value boundary values $0$ on the left boundary and $\log r'$ on the right boundary. As $it,\,t\in \mathbb{R}$, moves along $l$ with unit speed, we have $u'(it)=\partial_y u(it)$ and $v'(it)=\partial_y v(it)=i\partial_xu(it)$. It is not hard to see, using e.g. extremal distances, that $r-1\asymp r'-1\sim\log r'$ as $r\to 1$. We aim to show that the horizontal component of the movement converges uniformly to zero with all derivatives.
Assume towards contradiction that the desired uniform convergence does not hold. Then, for some $n$, there are sequences $r_k,t_k$ such that $|\frac{d^{(n)}}{d\tau^n} u(it_k)|>c>0,$ where $t(\tau)$ is a reparametrization such that $\frac{d}{d\tau} v(it_k)\asymp 1.$ (Or, there is a sequence such that $\frac{d}{d\tau} v(it_k)\leq 0$, which will likewise lead to a contradiction.) We may assume by passing to a subsequence that $t_k\to t_0$. The idea is to arrive at a contradiction by showing that as we zoom in near $t_k$, the function $u$, appropriately rescaled, converges to an affine function $\Re e\,z$.
Assume first $it_0\notin \log \partial D$. We "zoom in" to the scale $\log r_k$ around $t_k$, that is, put $u_k(z)=u((z\log r_k+it_k)-\log r'_k$; this is a sequence of negative harmonic functions, defined on an increasing sequence of domains exhausting the half-plane $\{\Re e\,z<1\}$ and vanishing on the boundary. We can write $u_k(0)$ using the Poisson kernel in the half-disc $B$ with center $1$ of large radius $R_k=c\log r_k$:
$$
u_k(z)=\int_{\partial B\cap{\Re e\, w<1}}P_{R_k,z,w}u(w)|dw|,
$$
where
$$P_{R,z,w}=\frac{1}{2\pi R}\Re e\,\left(\frac{z+(w-1)}{z-(w-1)}-\frac{z+(2-\bar{w})}{z-(2-\bar{w}) }\right).$$
For this, and taking into account that $|u_k|\leq \log r'_k\asymp \log r_k$, one can work out the asymptotics of $u_k$ and its derivatives, uniformly on compacts near $z=0:$
$$
u_k=c_k(\log r_k)^2\Re(z-1)+O(\log r_k)^3;
$$
where $c_k>0$ is uniformly bounded from above and below, and
$$
\partial^{n}_z u_k=c_k(\log r_k)^2\partial^n_z\Re(z-1)+O((\log r_k)^{3+n}).
$$
Reading off the asymptotics of the derivatives of $u$, we will arrive at a contradiction with the assumption.
In the case $t_0\in\partial \log D$, there are three subcases: either $\mathrm{dist}(t_k,\partial \log D)/\log r_k\to 0$, or $\mathrm{dist}(t_k,\partial \log D)/\log r_k\to 1$, or neither; in the third subcase we may assume, passing to a further subsequence, that the ratio tends to some limit $x$ between $0$ and $1$. The case $\mathrm{dist}(t_k,\partial \log D)/\log r_k\to 1$ is done as above. The case $\mathrm{dist}(t_k,\partial \log D)/\log r_k\to 0$ is similar, once you intertwine with a periodic conformal map $\phi:\mathbb{C}\setminus \log D\to{\Re z>0}.$ This map straightens the left boundary of \log D and does not depend on $r$, the line $\varphi(l)$ is now a $C^\infty$ curve touching the boundary, and the asymptotic analysis of $u$ as we zoom in near $u(\phi(it_k))$ is as above. In the final subcase, as we zoom in near $it_k$, the domain converges to a strip $-x<\Re z< 1-x$ (with a speed we have some control of), so the leading term in the asymptotics of $u_k$ will be again an affine function.
