Ok, I figured out the answer to my own question, though without the $\ln(ep K^2/\sigma^2)$ in the denominator. I'll see whether I can get that later (or maybe someone else sees it?). We can assume without loss of generality that $\mathbb{E} X_i = 0$ for each $i$ (otherwise apply what follows with the random variables replaced by $X_i - \mathbb{E} X_i$). The trick is to use gaussians and not signs, since then Khintchine's inequality is correct as both an upper and lower bound. Let $\sigma_i$ be i.i.d. Rademacher and $g_i$ be i.i.d. normal.
\begin{align}
\|\sum_i X_i\|_p &= \|\sum_i X_i - \mathbb{E} X_i\|_p\\
&\le \|\sum_i (X_i - X_i')\|_p\\
&\le 2\|\sum_i \sigma_i X_i\|_p\\
&= 2\sqrt{\frac{\pi}{2}}\|\mathbb{E}_g\sum_i \sigma_i |g_i| X_i\|_p\\
&\lesssim \|\sum_i \sigma_i |g_i| X_i\|_p\\
&= \|\sum_i g_i X_i\|_p (1)\\
&\le \sqrt{p} \|(\sum_i X_i^2)^{1/2}\|_p\\
&\le \sqrt{p} \|\sum_i X_i^2\|_p^{1/2}\\
&= \sqrt{p} \|\sum_i X_i^2 - \mathbb{E}\sum_i X_i^2 + \sum_i X_i^2\|_p^{1/2}\\
&\le \sigma\sqrt{p} + \sqrt{p}\|\sum_i X_i^2 - \mathbb{E}\sum_i X_i^2\|_p^{1/2}\\
&\lesssim \sigma\sqrt{p} + \sqrt{p}\|\sum_i g_i X_i^2\|_p^{1/2}\\
&= \sigma\sqrt{p} + \sqrt{p}\left(\mathbb{E}_X\|X\|_\infty^p\left(\mathbb{E}_g|\sum_i g_i X_i \frac{X_i}{\|X\|_\infty}|^p\right)\right)^{1/2p}\\
&\le \sigma\sqrt{p} + \sqrt{pK}\left(\mathbb{E}_X\mathbb{E}_g|\sum_i g_i X_i \frac{X_i}{\|X\|_\infty}|^p\right)^{1/2p}\\
&\lesssim \sigma\sqrt{p} + \sqrt{pK}\left(\mathbb{E}_X\mathbb{E}_g|\sum_i g_i X_i|^p\right)^{1/2p} (2)\\
&= \sigma\sqrt{p} + \sqrt{pK}\|\sum_i g_i X_i\|_p^{1/2} (3)
\end{align}

Inequality (2) used the gaussian property that Khintchine is not only an upper bound, but also a lower bound up to a constant independent of p. Now notice we have bounded (1) in terms of its square root in (3). Thus $\|\sum_i g_i X_i\|_p^{1/2}$ is at most the larger root of the associated quadratic equation, and this gives us the bound
$$
\|\sum_i g_i X_i\|_p \lesssim \sigma\sqrt{p} + Kp
$$
as desired.

I tried to be careful and write all the details to make it clear, but much of the above argument can be compressed once you're familiar with the right bag of tricks (symmetrization, gaussian comparison principle, and triangle inequality in the right places).