I contacted the author, and he was kind enough to write up a proof for me. I have attempted to simplify his proof for presentation here. I also use some notation without explanation to reduce clutter; the meanings should be clear. The trick is to use an apparently well known result from lattice theory.
Proposition 1 (Lattice Theory rotation trick)
If $b_1, b_2, \ldots, b_\ell$ are real numbers such that
\begin{align*}
b_1 + b_2 + \cdots + b_\ell = r \ell
\end{align*}
($\ell \geq 2$) then the lattice path
\begin{align*}
(0, 0), (1, b_1), (2, b_1 + b_2), \ldots, (\ell, b_1 + b_2 + \cdots + b_\ell) = (\ell, r \ell)
\end{align*}
has a cyclic forward shift by some $k$ with $0 \leq k \leq \ell - 1$
\begin{align*}
(0, 0), (1, b_{k+1}), (2, b_{k+1} + b_{k+2}), \ldots, (\ell, b_{k+1} + b_{k+2} + \cdots + b_\ell + b_1 + b_2 + \cdots + b_k) = (\ell, r \ell)
\end{align*}
so that
\begin{align*}
b_{\overline{k+1}} + b_{\overline{k+2}} + \cdots + b_{\overline{k+j}} \leq jr
\end{align*}
for all $1 \leq j \leq \ell$, where $\overline{k+i} \equiv k + i \pmod{\ell}$ and $1 \leq \overline{k+i} \leq \ell$.
Proof
Let $k$ be the smallest index such that the point $(k, b_1 + b_2 + \cdots + b_k)$ is not above the line $y = rx$ and the distance between this point and the line is maximum. Notice that $k \leq \ell - 1$ by the extreme value theorem.
Corollary 2
If $n_1 < n_2 < \ldots < n_k$ and $r = n_k/k$, then there is some $\hat{k}$ with $1 \leq \hat{k} \leq k - 1$ such that
\begin{align*}
n_{k-\hat{k}+1} - n_j \geq (k - \hat{k} + 1 - j) r
\end{align*}
for all $1 \leq j \leq k - \hat{k}$.
Proof
Apply Propositon 1 to the sequence
\begin{align*}
b_1 = n_k - n_{k-1}, b_2 = n_{k-1} - n_{k-2}, \ldots, b_{k-1} = n_2 - n_1, b_k = n_1
\end{align*}
Lemma 3 (bound on additive term)
For odd $x \in \mathbb{N}$, suppose that
\begin{align*}
T^n(x) = \dfrac{3^k}{2^n}x + e(x, k)\;\;\;\;\; e(x, k) = \sum_{i=0}^{k-1} \dfrac{3^i}{2^{n_k - n_{k-1-i}}}
\end{align*}
where $r = n/k \geq \log_2 3$. Then there is a $1 \leq \hat{k} < k$ such that
\begin{align*}
e(x,k-\hat{k}) \leq \dfrac{1}{2^r - 3}
\end{align*}
Proof
Let $r = n/k = (\log_2 3)(1 + \delta)$, where $\delta > 0$, and apply Corollary 2 to $0 = n_0 < n_1 < \cdots < n_{k-1}$ to find an index $\hat{k}$ such that $1 \leq \hat{k} < k$ and
\begin{align*}
n_{k-\hat{k}} - n_{j-1} \geq (k - \hat{k} - j + 1)r
\end{align*}
for all $1 \leq j \leq k - \hat{k}$. Then,
\begin{align*}
2^{n_{k-\hat{k}} - n_{i-1}} \geq 2^{(k-\hat{k}-i+1)r} = 3^{(k-\hat{k}-i+1)(1+\delta)}
\end{align*}
and
\begin{align*}
e(x, k - \hat{k}) & = \sum_{i=1}^{k-\hat{k}} \dfrac{3^{k-\hat{k}-i}}{2^{n_{k-\hat{k}}-n_{i-1}}} \\
& \leq \sum_{i=1}^{k-\hat{k}} \dfrac{3^{k-\hat{k}-i}}{3^{(k-\hat{k}-i+1)(1 + \delta)}} \\
& = \dfrac{1}{3} \sum_{i=1}^{k-\hat{k}} \dfrac{1}{3^{(k-\hat{k}-i+1)\delta}} \\
& \leq \dfrac{1}{3^{1+\delta} - 3} \\
& = \dfrac{1}{2^r - 3}
\end{align*}
Remark
The key fact about Lemma 3 is that the bound is independent of both $x$ and $e(x,k)$, depending only on the value of $r = n/k$. It achieves this by making the choice $\hat{k}$ that depends on $x$ and showing that such a choice must exist for every $x$ with $r \geq \log_2 3$. This is the ``trick."
Theorem 4
If $x, T(x), T^2(x), \ldots$ is a divergent trajectory with
\begin{align*}
k(x, n) = |\{T^j(x) \equiv 1 \pmod{2} : 0 \leq j < n\}|
\end{align*}
then
\begin{align*}
\liminf_{n \to \infty} \dfrac{k(x, n)}{n} \geq \log_3 2
\end{align*}
Proof
Since $x$ has a divergent trajectory, there must be a sequence $y_0 < y_1 < y_2 < \cdots$ such that $y_j = T^{n_j}(x)$, for some natural numbers $n_0 < n_1 < n_2 < \cdots$, and such that $T^n(y_j) > y_j$ for all $n \in \mathbb{N}$. For each (fixed) $y_j$, we have
\begin{align*}
\liminf_{n \to \infty} \dfrac{k(x, n)}{n} = \liminf_{n \to \infty} \dfrac{k(x, n) - k(x, n_j)}{n - n_j} = \liminf_{n \to \infty} \dfrac{k(y_j, n)}{n}
\end{align*}
Now, suppose that
\begin{align*}
\liminf_{n \to \infty} \dfrac{k(x, n)}{n} < \log_3 2
\end{align*}
Then, there is some constant $c$ such that
\begin{align*}
\dfrac{k(x, n)}{n} \leq \dfrac{1}{c} < \log_3 2
\end{align*}
infinitely often, and, in particular, for each $y_j$, there is always an $n$ such that
\begin{align*}
\dfrac{k(y_j, n)}{n} \leq \dfrac{1}{c} < \log_3 2
\end{align*}
Let $d = c - \log_2 3 > 0$. Then, $n \geq k(y_j, n)(\log_2 3 + d)$ implies
\begin{align*}
2^n \geq 3^{k(y_j, n)} 2^{kd} \geq 3^{k(y_j, n)} 2^d \iff \dfrac{3^{k(y_j, n)}}{2^n} \leq 2^{-d}
\end{align*}
Let $r = n/k(y_j, n) \geq c$. Applying Lemma 3, we have an $n^*$ such that
\begin{align*}
T^{n^*}(y_j) \leq 2^{-d} y_j + \dfrac{1}{2^r - 3} \leq 2^{-d} y_j + \dfrac{1}{2^c - 3}
\end{align*}
where the values $c$ and $d$ are constant across all the $y_j$ (i.e., for each $y_j$, there is an $n^*$ such that the bound holds). Since $2^{-d} < 1$ and $y_j$ grow unbounded, there is a
\begin{align*}
y_j > \dfrac{1}{(2^c - 3)(1 - 2^{-d})}
\end{align*}
at which point
\begin{align*}
T^{n^*}(y_j) < y_j
\end{align*}
contradicting our construction of the $y_j$ and proving
\begin{align*}
\liminf_{n \to \infty} \dfrac{k(x,n)}{n} \geq \log_3 2
\end{align*}
Remark
If you find any errors in the above, it is almost certainly from my attempt to simplify the proof I was given and not from the author.
