= Solution
Write $T_p=p^2/(2m)$ and $V=m\omega^2x^2/2$. Dropping <commutators> of order $\beta^2$ permits $e^{-\beta H}\simeq e^{-\beta V}e^{-\beta T_p}$. With $\langle x|p\rangle=e^{ipx/\hbar}/\sqrt{2\pi\hbar}$, the <momentum> Gaussian gives
$$
\langle x|e^{-\beta H}|x'\rangle\simeq
\sqrt{\frac{m}{2\pi\beta\hbar^2}}\exp\left[-\frac{m(x-x')^2}{2\beta\hbar^2}-\frac{\beta m\omega^2x^2}{2}\right].
$$
Integrating the diagonal yields $Z\simeq1/(\beta\hbar\omega)$, so the <leading high-temperature density kernel> is
$$
\boxed{\langle x|\rho|x'\rangle\simeq
\omega\sqrt{\frac{\beta m}{2\pi}}\exp\left[-\frac{m(x-x')^2}{2\beta\hbar^2}-\frac{\beta m\omega^2x^2}{2}\right]}.
$$
This unsymmetrized product has an apparent off-diagonal asymmetry at the omitted order; reversing the product puts $x'$ in the potential factor. A manifestly <Hermitian> approximation uses $e^{-\beta V/2}e^{-\beta T_p}e^{-\beta V/2}$, replacing $\beta m\omega^2x^2/2$ by $\beta m\omega^2(x^2+x'^2)/4$ with the same leading normalization. The condition is $\beta\hbar\omega\ll1$; the Gaussian off-diagonal width is of order $\hbar\sqrt{\beta/m}$.
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