Let $Q$ be a compact self-adjoint operator on $L^2(\mathbb R,\mathbb C)$. Notice that $(1-\Delta)^{s}$ is a positive (and hence self-adjoint) operator on $H$ for any $s>0$.
We denote by $\mathfrak S^p$ the Schatten space with norm $\|A\|_{\mathfrak S^p}=(\mathrm{Tr}|A|^p)^{\frac{1}{p}}$.
I am trying to show that, for $p>1$, $$\|(1-\Delta)^{\frac{s}{2}}Q(1-\Delta)^{\frac{s}{2}}\|_{\mathfrak S^p}\leq \|(1-\Delta)^{\frac{s}{2}}|Q|^{\frac{1}{2}}\|_{\mathfrak S^{2p}}^2.$$
Some essential ingredients would be that $\|A^*\|_{\mathfrak S^p}=\|A\|_{\mathfrak S^p}$ for all $p\geq 1$, and that Schatten norms satisfy the Hölder inequality $\|AB\|_{\mathfrak S^r}\leq \|A\|_{\mathfrak S^{p}}\|B\|_{\mathfrak S^q}$ if $1/r=1/p+1/q$.
Another fact which may be useful: since $Q$ is self-adjoint, $Q\leq |Q|$, and therefore $(1-\Delta)^{\frac{s}{2}}Q(1-\Delta)^{\frac{s}{2}}\leq (1-\Delta)^{\frac{s}{2}}|Q|(1-\Delta)^{\frac{s}{2}}$.
Let $J = \|(1-\Delta)^{\frac{s}{2}}Q\,(1-\Delta)^{\frac{s}{2}}\|_{\mathfrak S^p}$. The polar decomposition of $Q$ allows to write $Q = U \,|Q|$ for some operator $U$ with $U_{|\,\mathrm{Ker}|A|} = 0$ and $U_{|\,\overline{\mathrm{Ran}|A|}}$ is an isometry and which commutes with $|Q|$ (see e.g. here), that is $U\,|Q|\,U^* = |Q|$ and so taking the square root, $U\,|Q|^{1/2}\,U^* = |Q|^{1/2}$, so that $U\,|Q|^{1/2} = |Q|^{1/2}\,U$ (that is, $U$ also commutes with $|Q|^{1/2}$). Therefore, $Q = |Q|^{1/2}\,U\,|Q|^{1/2}$ which implies that $$ J = \|(1-\Delta)^{\frac{s}{2}}\,|Q|^{1/2}\,U\,|Q|^{1/2}\,(1-\Delta)^{\frac{s}{2}}\|_{\mathfrak S^p} $$ and so by Hölder's inequality and the fact that $\|U\|_{\mathfrak S^\infty} = 1$ $$ J \leq \|(1-\Delta)^{\frac{s}{2}}\,|Q|^{1/2}\|_{\mathfrak S^{2p}} \||Q|^{1/2}\,(1-\Delta)^{\frac{s}{2}}\|_{\mathfrak S^{2p}}. $$ This and the fact that the Schatten norm is invariant by taking the adjoint lead to $$ J \leq \|(1-\Delta)^{\frac{s}{2}}\,|Q|^{1/2}\|_{\mathfrak S^{2p}}^2. $$