Nik's answer seems perfectly suited to what the OP was looking for, but there are indeed other generalizations of complete regularity to C-algebras, as shown in Rosicky's "Multiplicative lattices and C-algebras" Cahiers Top. Geom. Diff. Cat. tome 30 no 2 (1989), 95-110. To paraphrase part of that paper, note first that complete regularity can be stated in terms of open and closed sets, without reference to points. Specifically, a topological space $X$ is completely regular iff, for every open $O\subseteq X$, $$O=\bigcup\{N:N\text{ is open and }\overline{N}\subseteq O\}.$$ So a natural question to ask is whether, for any C*-algebra $A$ and open projection $q\in A^{**}$, $$q=\bigvee\{p\in A^{**}:p\text{ is an open projection and }\overline{p}\leq q\}?$$ The answer is yes. To see this, take any $a\in A^1_+$ with $a\leq q$ and note that $\overline{a_{(\epsilon,1]}}\leq a_{[\epsilon,1]}\leq a_{(0,1]}\leq q$, for any $\epsilon>0$, where $a_S$ denotes the spectral projection of $a$ in $A^{**}$ corresponding to $S\subseteq\mathbb{R}$. The result now follows because $q$ being open means $$q=\bigvee\{a_{(0,1]}:a\in A^1_+\text{ and }a\leq q\}.\\\\$$
As $\overline{N\cup M}=\overline{N}\cup\overline{M}$, for open subsets $N$ and $M$, above we actually have a directed union. On the other hand, we can have $\overline{p\vee r}\neq\overline{p}\vee\overline{r}$ for open projections $p$ and $r$, as in Example II.6 in Akemann's paper referred to by Nik, which also shows that the supremum above is not always directed. However, we might ask, $$\text{can the open projections above be replaced by a directed subset?}$$ In general I do not know. If $A$ is separable then the answer is yes, as then every open $p$ will be of the form $a_{(0,1]}$, for some $a\in A^1_+$, and $a_{(\epsilon,1]}$ is directed, for $\epsilon>0$. More generally, the answer is yes if $A$ has an "approximately idempotent" approximate unit $(a_\lambda)\subseteq A^1_+$, i.e. such that $a_\lambda a_\gamma=a_\lambda$ whenever $\lambda\leq\gamma$. I do not even know of a C*-algebra that does not have such an approximate unit, but presumably they exist - perhaps Nik can help me out here?
Incidentally, note the relation $\overline{N}\subseteq O$ can be expressed purely in terms of open sets as $$\text{there exists open $M$ disjoint from $N$ such that }X=M\cup N.$$ Thus we might wonder if, for open projections $p,q\in A^{**}$, $$\overline{p}\leq q\quad\Leftrightarrow\quad p\leq q\text{ and there exists open $r\in A^{**}$ with $r\wedge p=0$ and $r\vee q=1$?}$$ Certainly $\Rightarrow$ holds although there are non-unital $A$ where $\Leftarrow$ does not. But perhaps equivalence holds for unital $A$?