I assume that your manifold and foliation are smooth and foliation is of codimension 1, otherwise see Jack Lee'a comment. Then pick a Riemannian metric on $X$ and at each point $x\in M$ take unit vector $u_x$ orthogonal to the leaf $F_x$ through $x$: There are two choices, but since your foliation is transversally orientable, you can make a consistent choice of $u_x$. Then $u$ is a nonvanishing vector field on $X$.
In fact, orientability is irrelevant: Clearly, it suffices to consider the case when $X$ is connected. Then you can pass to a 2-fold cover $\tilde{X}\to X$ so that the foliation ${\mathcal F}$ on $X$ lifts to a transversally oriented foliation on $\tilde{X}$. See Proposition 3.5.1 of
A. Candel, L. Conlon, "Foliations, I", Springer Verlag, 1999.
You should read this book (and, maybe, its sequel, "Foliations, II") if you want to learn more about foliations.
Then $\chi(\tilde{X})=0$. Thus, $\chi(X)=0$ too. Now, recall that a smooth compact connected manifold admits a nonvanishing vector field if and only if it has zero Euler characteristic. Thus, $X$ itself also admits a nonvanishing vector field.
Incidentally, Bill Thurston proved in 1976 (Annals of Mathematics) that the converse is also true: Zero Euler characteristic for a compact connected manifold implies existence of a smooth codimension 1 foliation. This converse is much harder.
This is true for any smooth circle action on a closed manifold. There is no need to have an almost complex structure.
In fact, more generally we have the following.
Suppose $S^1$ acts smoothly on a closed manifold $M$. Then the fixed point set $M^{S^1}$ has the property that $\chi(M) = \chi(M^{S^1})$.
(So, in particular, the result holds even if the fixed point set is not isolated).
Proof: By averaging an arbitrary Riemannian metric on $M$, we may assume the action is isometric. The fixed point set of an isometric action is always a totally geodesic submanifold, but it may have several components. (It can only have finitely many components since it is compact). In particular, asking about $\chi(M^{S^1})$ makes sense.
We also point out that in this situation, the number of isotropy groups is finite. (This is true, up to conjugacy, for any compact Lie group action on any closed manifold, as a consequence of the slice theorem).
Let $N$ be a component of $M^{S^1}$ and let $\nu N$ be an embedding of the normal bundle of $N$ into some $\epsilon$-neighborhood of $N$. By shrinking $\epsilon$, we may assume that $\nu N_1 \cap \nu N_2 = \emptyset$ for disjoint components $N_1,N_2\subseteq M^{S^1}$. I will use the notation $\nu M^{S^1}$ to denote the the union of the $\nu N_i$
Note that the $S^1$ action preseves $\nu M^{S^1}$ since is is characterized as the set of points a distance $< \epsilon$ away from $M^{S^1}$ and the action is isometric.
It follows that the $S^1$ action also preserves $M\setminus \nu M^{S^1}$. Since we've removed the points with isotropy group $S^1$, and all other closed subgroups of $S^1$ are finite (and there are only finitely many of them) there is a neighborhood $U$ of the identity $1\in S^1$ with the property that any $p\in U$ acts with no fixed points. In particular, the action field ( $\frac{d}{dt}|_{t=0} e^{it} \cdot m$, for $m\in M$) has no zeros. By Poincare-Hopf, $M\setminus \nu M^{S^1}$ has zero Euler characteristic.
In a similar fashion, the $S^1$ action preserves the boundary $\partial \nu M^{S^1}$ since $\partial \nu M^{S^1}$ consists of all points in $M$ a distance $\epsilon$ from $N$. Repeating the argument in the previous paragraph, we deduce $\partial \nu N$ also has zero Euler characteristic.
Now we are basically done. Write $M = (M\setminus \nu M^{S^1}) \cup \nu M^{S^1}$. Using the fact that each $\nu N_i$ deformation retracts to $N_i$ (so, in particular, $\chi(\nu M^{S^1}) = \chi(M^{S^1})$, we compute \begin{align*} \chi(M) &= \chi(M\setminus \nu M^{S^1}) + \chi(\nu M^{S^1}) - \chi(\partial \nu M^{S^1})\\&= 0 + \chi(\nu M^{S^1}) + 0 \\ &= \chi(M^{S^1}). \end{align*}
Best Answer
Yes. In fact, every $\mathbb{R}P^{2n}$ admits a (smooth) circle action with exactly one fixed point. You can find a proof (as well as an extended discussion of related problems) in Circle Action with a Prescribed Number of Fixed Points by Ping Li.
For completeness, here's an excerpt from Example $2.1$ in that paper:
Note the author uses superscripts to indicate the real dimension of a space, so $\mathbb{C}^{2n}$ is the complex space with $2n$ real dimensions that we would normally call $\mathbb{C}^n$.
Your particular example $\mathbb{C}P^2 \# T^4$ is orientable, which means it cannot admit an action with exactly $1$ fixed point. In Circle Actions on Oriented Manifolds with Discrete Fixed Point Sets and Classification in Dimension $4$ by Donghoon Jang, we see (on page $2$):
I̶ ̶d̶o̶n̶'̶t̶ ̶h̶a̶v̶e̶ ̶a̶ ̶p̶r̶o̶o̶f̶ ̶o̶f̶ ̶t̶h̶i̶s̶ ̶c̶l̶a̶i̶m̶ ̶o̶n̶ ̶h̶a̶n̶d̶, but it means you'll need to consider nonorientable manifolds to get what you want.
Edit: I found a proof in a different paper by Donghoon Jang, Circle Actions on Oriented Manifolds with Few Fixed Points. This is part of Lemma $2.1$, which crucially uses the Atiyah-Singer Index Theorem.
I hope this helps ^_^