A Singular Foliation of $\mathbb{C}P^2$ which does not admit a global transverse submanifold Is there  a  singular  holomorphic foliation $F$ of  $\mathbb{C}P^2$ which  does not  admit  a global  transverse holomorphic  curve?  More  precisely there is no  an  immersed holomorphic submanifold  of  $\mathbb{C}P^2$  which intersect  all  regular  leaves, transversely? If there  exist  such  an  example $F$, does this  foliation admit  a smooth(but  not  necessarily holomorphic) global  transverse submanifold (of  real  dimension 2)?
 A: The answer to this question is negative. If we require the transversal surface to be an immersed holomorphic curve then the only foliation for which such a surface exists is the pencil of lines. This is proven the Claim below. I will consider more generally the second part of your question where we require the transversal surface to be smooth but not necessarily holomorphic.
I will produce examples of foliations that don't have such a smooth transversal surface at the end of this answer. And I guess in reality that the only foliation that possesses a transversal surface is the pencil of lines (though I can not prove it so far).
Note that each singular holomorphic foliation on $\mathbb CP^2$ defines a rank one subsheaf of $T\mathbb CP^2$ and its first Chern class is at most $1$. 
I claim  the following statement concerning transversal pairs $(\cal F,S)$ where $S$ is smooth (but not necessarily holomorphic). The case when $S$ is holomorphic immersed is proven in the same way.
Claim. There exists at most two types of transversal pairs $(S,\cal F)$ with $S$ smooth. 
1)The first type is when $\cal F$ is a pencil of lines and $S$ is a sphere of degree $1$. 
2) The second type is when $S$ is a torus in $\mathbb CP^2$ of degree $0$. Let us first prove this claim, first giving a remark. In particular $S$ can not be holomorphic.
Remark. Is $S$ is transversal to $\cal F$, it is orientable. Moreover we can choose its orientation in such a way, that it intersects the leafs of $\cal F$ positively.
Proof of Claim. Let $\cal F$ be a singular holomorphic foliation and let $S$ be a surface transversal to $\cal F$. Let $L$ be the line sub-bundle of the restriction $T\mathbb CP^2$ to $S$ such that $L$ is tangent to $\cal F$ along $S$.
Let us choose an orientation on $S$ so that the homology class of $S$ in $H_2(\mathbb CP^2)$ is $d\ge 0$. Note $S$ is orientable, because it is transversal to $\cal F$). Then since $\cal F$ is transversal to $S$, over $S$ $T\mathbb CP^2|_S$ splits into the sum  $L\oplus TS$, so we have  
$$3d=c_1(T\mathbb CP^2|_S)=c_1(L\oplus TS)=c_1(L)+2-2g=dc_1({\cal F})+2-2g\;\;\;\;(1)$$
$$d(3-c_1({\cal F}))=2(1-g).$$
Since $c_1({\cal F})\le 1$ we conclude that only two possibilities can occur  
1) $d=1$, $g=0$, and $c_1({\cal F})=1$. 
2) $d=0$ and $g=1$.
$\square$
Now, I don't know if examples of second type exist, when $S$ is a torus and $deg(S)=0$. It is hard to believe that they exist. But at least let me give some examples of foliations of degree $0$ that don't have a transversal surface.
Example. Let $\cal F$ be a foliation on $\mathbb CP^2$ tangent to the orbits of action of some $\mathbb C^*$ action on $\mathbb CP^2$. This foliation has degree $0\ne 1$, so by the above claim the only surface transversal to $\cal F$ can be a degree $0$ torus. I claim that such tori don't exist. 
Indeed, suppose by contradiction that $S$ is such a torus. Let's take a point $x\in S$. Then there is a $\mathbb C^*$-orbit $O$ passing through $x$. The closure of $O$ is a complex curve $\overline O$ in $\mathbb CP^2$. By the Remark in the beginning, there is an orientation of $S$ such that all points of intersection of $S$ with $\overline O$ are positive, i.e. $S\cdot \overline O\ge 1$. This contradicts to the fact that $\deg(S)=0$. $\square$.
