$\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}}
\newcommand{\PP}{\operatorname{\mathsf P}}$
In the excellent answer, Willie Wong offered a construction of a measure with a lacunary support set, disproving the conjecture. Let me offer another construction, where the support of the measure is a continuum; however, the construction is still lacunary in the sense that it involves sequences diverging faster than any exponential sequence.
The main idea is this. Let $p>0$ and $s>0$. Let $\|f\|_p:=\|f\|_{L^p(\mu)}$. Let $f(x)=x$ for all real $x$. Let us start with this trivial lemma:
Lemma 1. If $\mu(dx)=\frac1s\,e^{-x/s}I\{x>0\}$, where $I$ denotes the indicator, then
\begin{equation*}
\|f\|_p^p=\int_0^\infty x^p \,e^{-x/s}\frac{dx}s=s^p\Ga(p+1).
\end{equation*}
Then, by Stirling's formula, $\ln\|f\|_p\sim \ln(ps)$ as $s,p\to\infty$. So, if we alternate $s$ between $p$ and $p^2$, $\ln\|f\|_p/\ln p$ will alternate between $2$ and $3$. Now we need to glue pieces of such two alternating sequences of measures, with $s$ alternating between $p$ and $p^2$, to get one measure. The guiding idea in doing this is that for large $p$ the mass $m_p(dx):=x^p \,e^{-x/s}I\{x>0\}\frac{dx}s$ is mostly concentrated near the point $ps$.
To proceed, we shall need two more simple lemmas, which will be proved at the end of this answer.
Lemma 2. For every $y\in[0,ps]$ there is some $c\in[1/2,1]$ such that
\begin{equation*}
\int_y^\infty x^p \,e^{-x/s}\frac{dx}s=cs^p\Ga(p+1).
\end{equation*}
Lemma 3. For every $y\ge2ps$,
\begin{equation*}
\int_y^\infty x^p \,e^{-x/s}\frac{dx}s
\le\exp\{p\ln(2ps)-y/(2s)\}.
\end{equation*}
Let
\begin{equation*}
\mu(dx)=\sum_{j=1}^\infty\frac{dx}{s_j}\,e^{-x/s_j}I\{x_j<x<x_{j+1}\},
\end{equation*}
where
\begin{equation*}
s_j:=
\begin{cases}
p_j&\text{ if $j$ is odd}, \\
p_j^2&\text{ if $j$ is even},
\end{cases}
\qquad
p_j:=2^{2^j},\qquad x_j:=p_j s_j,
\end{equation*}
so that $p_{j+1}=p_j^2$.
Then
\begin{multline*}
\|f\|_p^p=\int_\R x^p\mu(dx)=\sum_{j=1}^\infty I_j(p),\\
I_j(p):=\int_{x_j}^{x_{j+1}} \exp\{g_{j,p}(x)\}\frac{dx}{s_j},\\
g_{j,p}(x):=p\ln x-x/s_j.
\end{multline*}
In view of Lemma 1 and Stirling's formula, for large odd $k$ one has
\begin{equation*}
I_k(p_k) \le s_k^{p_k} p_k^{p_k}=p_k^{2p_k} \tag{1}
\end{equation*}
and, for $j<k$,
\begin{equation*}
I_j(p_k)\le s_j^{p_k} p_k^{p_k} \le p_j^{2p_k} p_k^{p_k}
\le p_k^{2p_k},
\end{equation*}
whence
\begin{equation*}
\sum_{j<k}I_j(p_k) \le k p_k^{2p_k}=p_k^{(2+o(1))p_k}. \tag{2}
\end{equation*}
Next, for large odd $k$ and $j>k$,
\begin{equation*}
p_k\ln(2p_ks_j)-p_k\ln(p_ks_k)
\le p_k\ln s_j\le p_k\ln(p_j^2)\le p_j^{1/2}\ln(p_j^2)\le p_j/4,
\end{equation*}
whence, by Lemma 3,
\begin{multline*}
I_j(p_k)\le\exp\{p_k\ln(2p_ks_j)-p_j/2\}\le\exp\{p_k\ln(p_ks_k)-p_j/4\} \\
\le2^{-(j-k)}\exp\{p_k\ln(p_ks_k)\},
\end{multline*}
so that
\begin{equation*}
\sum_{j>k}I_j(p_k) \le p_k^{2p_k}. \tag{3}
\end{equation*}
Collecting (1), (2), (3), for large odd $k$ we have
\begin{equation*}
\ln\|f\|_{p_k}\lesssim 2\ln p_k.
\end{equation*}
On the other hand, for large even $k$, by Lemma 2 and Stirling's formula,
\begin{equation*}
\ln\|f\|_{p_k}\ge\frac1{p_k}\,\ln I_k(p_k)= \ln(s_k p_k^{1+o(1)})
= \ln(p_k^{2} p_k^{1+o(1)})
\sim3\ln p_k.
\end{equation*}
So, $\ln\|f\|_p/\ln p$ does not converge as $p\to\infty$.
In conclusion, let us prove the lemmas.
Proof of Lemma 1. This is obvious.
Proof of Lemma 2. Write
\begin{equation}
x^p \,e^{-x/s}=e^{g(x)},\quad g(x):=p\ln x-x/s. \tag{4}
\end{equation}
Then $g'(ps)=0$ and $g''(x)$ is increasing in $x>0$. So, $g(ps-u)<g(ps+u)$ for $u\in(0,ps)$. So, for every $y\in[0,ps]$,
\begin{equation*}
\int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s
\ge\int_y^\infty x^p \,e^{-x/s}\frac{dx}s
\ge\frac12\,\int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s.
\end{equation*}
To complete the proof of Lemma 2, it remains to refer to Lemma 1.
Proof of Lemma 3. For $g$ as in (4) and for $x>2ps$, we have $g'(x)=\frac px-\frac1s\le-\frac1{2s}$ and hence
\begin{equation}
g(x)\le g(2ps)-\frac1{2s}(x-2ps)=p\ln(2ps)-\frac x{2s}.
\end{equation}
Now Lemma 3 easily follows.
