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The implication is false. Take the thin rectanglesEach is a compact H-hull, and $\mathrm{diam}(A_{\epsilon })=\sqrt{4+{\epsilon }^2}\to 2$. We verify that its half-plane capacity nevertheless tends to zero.

Let $K=\{z\in \mathbb{H}:|z|\le 2\}$, which contains all these rectangles. At the Brownian exit time $\tau$ from $\mathbb{H}\setminus A_{\epsilon }$, the exit height is at most $\epsilon$, and a positive exit height requires hitting $K$ before $\mathbb{R}$. ThereforeFor $y>2$, conformal invariance of planar Brownian motion and $g_K(z)=z+4/z$ identify the probability on the right with the harmonic measure of $[-4,4]$ from $i(y-4/y)$ in $\mathbb{H}$. Integrating the Poisson kernel for the upper half-plane givesThe Brownian representation of half-plane capacity now proves the explicit estimateChoose $\epsilon =1/n$ to obtain the required sequence. This is an instance of half-plane capacity of a low rectangle: shrinking height can make half-plane capacity small while horizontal extent stays fixed.

False. LetAfter filling bounded complementary components if necessary, this is a compact H-hull with diameter asymptotic to $2n$. The half-plane capacity of a low rectangle estimate, together with scaling and translation of half-plane capacity, givesThus unbounded diameter need not force unbounded capacity.

Write $A_r=[-r,r]\times (0,1]$ and take $r\ge 1$. By the Brownian representation of half-plane capacity,On hitting $A_r$, the exit height is at most one. Moreover, $A_r$ lies in the half-disc of radius $\sqrt{r^2+1}\le 2r$. The harmonic measure of that semicircle as viewed from $iy$ is $O(r/y)$: mapping its exterior to $\mathbb{H}$ by $z\mapsto z+(2r{)}^2/z$ reduces the estimate to the Poisson kernel for the upper half-plane on an interval of length $O(r)$. Consequentlyfor large $y$, and $\mathrm{hcap}(A_r)\le Cr$. This is the half-plane capacity of a low rectangle.

For $r\ge 1$, let $R_r=[-r,r]\times (0,1]$. The Brownian representation of half-plane capacity givesOn hitting $R_r$ the imaginary part is at most one, while the harmonic measure estimate supplied in the question shows that the probability of reaching a disc of radius $O(r)$ containing $R_r$ is $O(r/y)$. Hence $\mathrm{hcap}(R_r)\le Cr$, the half-plane capacity of a low rectangle bound.

SetThe scaling and translation of half-plane capacity giveswhereas $\mathrm{diam}(A_n)=\sqrt{4+n^{-2}}\to 2$. This supplies the required sequence.