Let $G$ be a locally compact (Hausdorff) group and $H$ a Hilbert space. Given $\xi,\eta\in C_c(G)$, we denote by $\kappa_{\xi,\eta}$ the normal (i.e. ultraweakly continuous) linear functional on $B(L^2(G))$ given by $T\mapsto \langle T\xi,\eta\rangle$ and let $\kappa_{\xi,\eta}\overline{\otimes}\text{id}_{B(H)}\colon B(L^2(G))\overline{\otimes}B(H)\to B(H)$ be the associated ultraweakly continuous and completely positive slice map.
I want to conclude the following for an element $x\in B(L^2(G))\overline{\otimes}B(H)$: $$x=0 \Leftrightarrow \kappa_{\xi,\eta}\overline{\otimes}\text{id}_{B(H)}(x)=0 \quad \text{for every} \ \xi,\eta\in C_c(G) $$
Apparently, this should be "clear enough". All hints are appreciated.
Let $s,t \in H$ and note that $$0 = \kappa_{s,t}((\kappa_{\xi, \eta}\overline{\otimes} \operatorname{id})(x))= (\kappa_{\xi, \eta}\overline{\otimes}\kappa_{s,t})(x)= \kappa_{\xi \otimes s, \eta \otimes t}(x)= \langle x(\xi\otimes s), \eta \otimes t\rangle.$$ Since $C_c(G)$ is norm-dense in $L^2(G)$, it easily follows that $\langle x\mu, \mu'\rangle = 0$ for all $\mu, \mu' \in L^2(G)\otimes H.$ This implies $x=0$.