Let $K=Q(\sqrt{3})$ , is $SU(2,1)(K)$ dense in $SU(2,1)(C)$ for the complex topology?
Yes, the density you indicate certainly occurs. For $SU(n,n1)$ and $SO(n,n1)$ and many other classical groups, an analogous density holds, by a similar argument, although there are some minor complications (about compact real Lie groups...) for $SU(p,q)$ with $p>q+1$.
In the case at hand, first note that whatever the rational form of the hermitian form, there is a nontrivial isotropic vector, by Hasse's principle: the corresponding quadratic form is 6dimensional, so there are isotropic vectors padically everywhere locally. Then at the archimedean place the assumption on signature gives an isotropic vector. Thus, by HasseMinkowski, there is a rational isotropic vector.
The latter point ensures that we can (rationally) change coordinates to $\pmatrix{0 & 0 & \sqrt{3}\cr 0 & 1 & 0 \cr \sqrt{3} & 0 & 0}$ or other antidiagonal versions. In these coordinates, we can identify the (rational) parabolic subgroup $P=\pmatrix{a & x & y \cr 0 & \mu & x' \cr 0 & 0 & a^{1}}$, whose rational points are where $a\in {\mathbb Q}^\times$, $x\in K$, and $y$ and $x'$ are determined by $x$ and $a$, and $\mu=1$. The rational points are dense in the real points of $P$. The only slightly tricky part is noting that elements $\mu=1$ of $K$ are dense in the unit circle, but this follows from an argument similar to the parametrization of Pythagorean triples. This is the step that becomes less elementary when the corresponding compact part of the minimal rational parabolic is larger.
Then verify the Bruhat decomposition, that the rational points of the whole group are the union $P\cup PwP$, where $w$ is the "long Weyl group element" antidiagonal $1,1,1$. The big cell $PwP$ in this case is readily identifiable as elements of the group (with the antidiagonal form as above) with nonzero lower left entry.
For bigger unitary and orthogonal groups, there are more Weyl group elements, and proving the Bruhat decomposition is less trivial, but this densityofrationalpoints argument does not need the sharpest form of it.
There are also general results (due to A. Borel) that linear reductive groups are "unirational", so density of rational points holds quite generally, for a general reason. Nevertheless, it is also verifiable (via Bruhat, etc., as above) casebycase.
Edit: It is probably worth being more careful in parsing this question, as observed in comments by @mikhail and @yves... First, $SU(p,q)$'s are not "complex" Lie groups, but real, even though the usual model uses complex numbers in the construction: complex conjugation is not complexlinear. Yes, the "strong" topology does arise as the subspace topology from the ambient complex space, in that model. Similarly, $SU(2,1)(K)$ is not really the $K$rational points, but $\mathbb Q$rational. For a commutative $\mathbb Q$algebra $A$, the $A$points are $SU(2,1)(K\otimes_{\mathbb Q}A)$. Thus, the complex points (in this more intrinsic sense) are isomorphic to $SL(3,\mathbb C)$.
Further, indeed, as @mikhail notes, the instantaneous conclusion that rational points are dense in the "strong" ("classical"?) topology does not follow so glibly from the unirationality per se, since all that truly guarantees (in general) is density in the much coarser Zariski topology. The refinement necessary to make this sort of argument work sometimes more generally, as in @yves' remark, perhaps makes it less persuasive as a causal mechanism, but it suggests that we are "close", in any case.

1$\begingroup$ @paul: How do you deduce that rational points are dense in the usual (complex) topology of $G$ from the fact that $G$ is unirational? Borel mentions Zariskidensity of rational points. Note that Sansuc constructed a semisimple group $G$ over $\mathbf{Q}$ such that $G(\mathbf{Q})$ is not dense in $G(\mathbf{Q}_2)$, see digizeitschriften.de/main/dms/img/?IDDOC=505419, Example 5.8. $\endgroup$ – Mikhail Borovoi May 3 '12 at 18:18

1$\begingroup$ @Mikhail: you're right. But since $G$ is simply connected and isotropic (it's geometrically isomorphic to $SL_3$), it's true that $G(\mathbf{R})=SU(2,1)(\mathbf{C})$ is generated by unipotents (an easier, less algebraic way to see this is just that $SU(2,1)(\mathbf{C})$ is connected for the ordinary topology) so the unirationality argument works. $\endgroup$ – YCor May 3 '12 at 18:53
(EDITED taking in account the comment of Yves.) The group denoted in the question by $SU(2,1)(K)$ is the group of $\mathbf{Q}$points $G(\mathbf{Q})$ for a suitable $\mathbf{Q}$group $G$, and the group denoted by $SU(2,1)(\mathbf{C})$ is $G(\mathbf{R})$. For any connected linear algebraic group $G$ over $\mathbf{Q}$, the group $G(\mathbf{Q})$ is dense in $G(\mathbf{R})$ for the real topology. This is the real approximation theorem for connected linear algebraic groups, see the reference to Sansuc's paper in my comments to A question on algebraic torus.
Note also that if $G$ is a simply connected semisimple group over $\mathbf{Q}$ (as the group in the question), then $G(\mathbf{Q})$ is dense in $G(\mathbf{Q}_p)$ for any prime $p$. This follows from the weak approximation theorem for simply connected groups (due to Kneser, Harder, Platonov). This is not true for semisimple groups that are not simply connected, see example 5.8 in Sansuc's paper, where a certain semisimple $\mathbf{Q}$group $G$ is constructed for which $G(\mathbf{Q})$ is not dense in $G(\mathbf{Q}_2)$.

$\begingroup$ What do you denote by $G(\mathbf{C})$? The group $SU(2,1)(\mathbf{C})$ is not the group of complex points of any algebraic group, but the group of real points. $\endgroup$ – YCor May 3 '12 at 18:55

$\begingroup$ @Yves: Thank you, I have corrected my stupid mistake. $\endgroup$ – Mikhail Borovoi May 3 '12 at 19:51