The answer to the second question is also negative. Let $(u_g)_{g\in G}$ be the generating unitaries of $C^{\ast}(G)$. Suppose that $(\varphi_i)_{i\in I}$ is a net a functions whose associated multipliers $m_{\varphi_{i}}: C^{\ast}_{r}(G) \to C^{\ast}_r(G)$ give CBAP of $C^{\ast}_{r}(G)$. Suppose now that the assignments $u_g \mapsto \varphi_i(g) u_g$ extend to uniformly completely bounded multipliers on the full $C^{\ast}$-algebra $C^{\ast}(G)$. It would follow that $C^{\ast}(G)$ has the CBAP because, by uniform boundedness, it would suffice to check convergence on finite sums $\sum a_g u_g$, which are dense in $C^{\ast}(G)$, and this is easy.

Let me now discuss, where the subtlety lies. The multipliers giving CBAP for non-amenable groups are not uniformly bounded in $B(G)$, the Fourier-Stieltjes algebra, and therefore do not give nice multipliers on $C^{\ast}(G)$. Indeed, for a function $\varphi: G \to \mathbb{C}$ with $\|\varphi\|_{B(G)}\leqslant C$ we get an associated multiplier $m_{\varphi}:C^{\ast}(G) \to C^{\ast}(G)$ via composition $C^{\ast}(G) \stackrel{\Delta}{\rightarrow} C^{\ast}(G) \otimes C^{\ast}(G) \stackrel{id \otimes \varphi}{\rightarrow} C^{\ast}(G)$, where $\Delta$ is a $\ast$-homomorphism such that $\Delta(u_g)=u_g\otimes u_g$, given by the universal property, and $\varphi:C^{\ast}(G)\to \mathbb{C}$ is the functional given by $\varphi$. Since $\ast$-homomorphisms are completely contractive, we get $\|m_{\varphi}\|_{cb}\leqslant C$.

The problematic behaviour of multipliers manifests itself in the fact that they are not *decomposable* as maps from $C^{\ast}_r(G)$ to itself. What I mean by this is that if we decompose them as combinations of positive multipliers, then the norms of the terms involved are not controlled by the cb norm of the multiplier itself. A positive multiplier on $C^{\ast}_r(G)$ necessarily comes from a positive definite function on $G$, hence automatically extends to a multiplier on $C^{\ast}(G)$, so I think that decomposability is the real issue here.