topology of $\mathbb{P}_2 (\mathbb{C}) \setminus \mathbb{P}_2 (\mathbb{R})$?

$H_i (CP^2 \setminus RP^2)\cong H^{4-i}(CP^2, RP^2)$ by Poincare-Alexander-Lefschetz duality (Bredon, Topology and Geometry, Theorem 8.3 on p. 351). The latter can be computed using the long exact sequence. $H^4 = Z$, $H^0=H^1=0$ is immediate. The piece

$$0 \to H^2 (CP^2,RP^2) \to H^2 (CP^2) \to H^2 (RP^2) \to H^3 (CP^2 , RP^2) \to 0$$

needs an extra argument. The map $Z=H^2 (CP^2 ) \to H^2 (RP^2)=Z/2$ is onto because the tautological complex line bundle restricts to the complexification of the real tautological line bundle, whose first Chern class generates $H^2 (RP^2)$. Thus $H_2 (CP^2 \setminus RP^2)=Z$ and $H_1 (CP^2 \setminus RP^2)=0$.

Because the first map in the above sequence is multiplication by $2$ (after identification with $Z$), it follows that the inclusion $H_2 (CP^2 \setminus RP^2)\to H_2 (CP^2)$ takes a generator to twice a generator. A generator of $H_2 (CP^2-RP^2)$ can be represented by an embedded sphere as follows. Take a quadric $Q \subset CP^2$ without real point, for example the one defined by the homogeneous equation $z_{0}^{2}+z_{1}^{2}+z_{2}^{2}=0$. By the degree genus formula, $Q$ has genus $0$, hence is a sphere. It lies in $CP^2 - RP^2$, and because its fundamental class is twice a generator of $H_2 (CP^2)$, it must represent a generator of $H_2 (CP^2 - RP^2)$.

In fact, $CP^2-RP^2$ is diffeomorphic to the normal bundle of $Q$, see Tom's comment below.

There is a beautiful (and elementary) paper by V. I. Arnold, which discusses this and generalizations.