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, there are other interesting questions based on how we interpret $\overline{N}\subseteq O$ in C*-algebras.  Specifically, 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 the following relations are equivalent for open projections $p,q\in A^{**}$,

 1. $\overline{p}\leq q$.
 2. $p\leq q$ and there exists open $r\in A^{**}$ with $rp=0$ and $r\vee q=1.$
 3. $p\leq q$ and there exists open $r\in A^{**}$ with $r\wedge p=0$ and $r\vee q=1.$

If $p$ is is compact then, using Akemann's non-commutative Urysohn lemma, from 1. we get $a\in A^1_+$ with $p\leq a\leq q$, which means we get 2. by taking $r=a_{[0,\epsilon)}$, for any $\epsilon\in(0,1)$.  In particular, 1.$\Rightarrow$2. whenever $A$ is unital, or even $\sigma$-unital, as we can then still get $a\in\mathcal{M}(A)^1_+$ with $p\leq a\leq q$, where $\mathcal{M}(A)$ is the multiplier algebra of $A$ - see p936 Lemma 3.31 of Brown's "Semicontinuity and multipliers of C*-algebras" Can. J. Math., Vol. XL, No. 4 (1988) 865-988.  Clearly 2.$\Rightarrow$3. although I can think of non-unital $A$ where the converse fails.  But as far as I know these could all be equivalent for unital $A$.  Does anyone else have more information on this?