I think this observation may lead to a solution. Notice that $\mathcal{A} \subset L^2(\mathcal{A})$ is dense and that
If it holds that, given $S \subset \mathcal{A} \subset L^2(\mathcal{A})$ a finite dimensional subspace, its orthogonal complement $$S^\perp = \{ \xi \in L_2(\mathcal{A}) : \phi(a^\ast \, \xi) = 0, \forall a \in S \}$$ has dense intersection with $\mathcal{A}$. Then (3) holds.
If for all $S \subset \mathcal{A}$, $S^\perp$ has dense intersection with the span of $U(\mathcal{A})$, the unitary group of $\mathcal{A}$, then (1) holds.
Observations 1 and 2 hold because they allow us to iteratively choose an orthogonal base with the desired properties.
I do not know whether 1 or 2 hold in general. Maybe you can try to use the fact that $\mathcal{A}$ is closed by functional calculus to define the functional $T_{a,b}: C_b(\mathbb{C}) \to \mathbb{C}$ by $$ f \mapsto \phi(a^\ast \, f(b)). $$ At least for normal $b$ the formula above makes sense. Then, the kernel of $T_{a,b}$ will have codimension 1 and picking an element $f$ in $$ \bigcap_{a \in S} \ker(T_{a,b}) $$ Will give you $f(b) \in S^\perp \cap \mathcal{A}$. Probably refining such type of argument you can prove observation 1 above.