Faithful traces on quasi-diagonal C*-algebras

Recall that a separable C*-algebra $$A$$ is quasi-diagonal if there are completely positive and contractive maps $$\varphi_k \colon A \rightarrow M_{n(k)}$$ such that $$||\varphi_k(ab) - \varphi_k(a)\varphi_k(b)|| \rightarrow 0$$ and $$||\varphi_k(a)|| \rightarrow ||a||$$ for every $$a, b \in A$$, where $$M_{n(k)}$$ denotes the algebra of $$n(k) \times n(k)$$ complex matrices.

The following ought to be known to experts, but I've had some issues trying to prove it.

Question: Let $$A$$ be a quasi-diagonal separable C*-algebra. Does $$A$$ admit a faithful trace?

Observe we do not assume that $$A$$ is simple. The idea would be to use an ultralimit of the cp maps $$\varphi_k \colon A \rightarrow M_{n(k)}$$ composed with the usual traces $$tr_{n(k)} \colon M_{n(k)} \rightarrow \mathbb{C}$$. This does not necessarily yield a faithful trace, though, as the traces $$tr_{n(k)}(\varphi_k(a^*a))$$ might be very small while $$||\varphi_k(a^*a)||$$ remains large. However, one could in some cases tweak $$\varphi_k$$ by adding a factor that enlarges the trace, maintains the asymptotic multiplicativity and the norm. The actual proof requires to do this everywhere at once, that is, for every possible element $$a^*a$$, and this seems impossible (or unreasonably hard).

Note, moreover, that there is no obvious K-theoretic obstruction, as every countable abelian group $$K_0(A)$$ embeds into $$\mathbb{R}$$.

Any help is greatly appreciated.

No, separable (unital) quasi-diagonal $$C^\ast$$-algebras do not necessarily admit a faithful tracial state. For instance, the $$C^\ast$$-algebra $$$$A= \{ f\in C([0,1], \mathcal O_2) : f(0) \in \mathbb C 1_{\mathcal O_2}\}$$$$ (where $$\mathcal O_2$$ is the Cuntz algebra with two canonical generators) is homotopic to $$\mathbb C$$, and hence is quasi-diagonal by the homotopy invariance of quasi-diagonality due to Voiculescu. However, any trace on $$A$$ vanishes on the ideal $$C_0((0,1], \mathcal O_2)$$, since this is purely infinite in the sense of Kirchhberg-Rørdam [Kirchberg, Eberhard; Rørdam, Mikael Non-simple purely infinite C∗-algebras. Amer. J. Math. 122 (2000), no. 3, 637–666.]. So the only tracial state on $$A$$ is the one factoring through evaluation at 0.
If you consider non-unital $$C^\ast$$-algebras, $$C_0((0,1], \mathcal O_2)$$ is an example of a separable quasi-diagonal $$C^\ast$$-algebra with no tracial states.
On the other hand, if you consider separable residually finite-dimensional (RFD) $$C^\ast$$-algebras, then they always admit a faithful tracial state: a separable $$C^\ast$$-algebra is RFD if it embeds into $$\prod_{n\in \mathbb N} M_{k(n)}(\mathbb C)$$ for some sequence $$k(n)$$ of natural numbers. As $$\prod_{n\in \mathbb N} M_{k(n)}(\mathbb C)$$ has a faithful tracial state, e.g. $$\sum_{n\in \mathbb N} 2^{-n} \mathrm{tr}_{k(n)}$$, so does any $$C^\ast$$-subalgebra.