Let $F$ be a compact oriented surface with a foliation $\cal F$ with $k$-prong singularities only (or, if it helps, assume that $\cal F$ admits an invariant measure). Is it true then there exists a pseudo-Anosov $\phi$ such that for every loop or an arc (with endpoints in $\partial F$) $\gamma$, $\phi^n(\gamma)$ is transversal to $\cal F$ for large $n$?


Here is a rough argument why it is probably true: The $MCG(F)$ action on ${\cal PMF}(F)$ is minimal, cf. Fathi-Laudenbach-Poenaru, Thurston's Work on Surfaces, for closed $F$. (A reference for bounded $F$ would be useful although the original proof seems to work). This implies that the unstable foliations $\cal F^u\_\phi$ of pseudo-Anosovs $\phi$ are dense in ${\cal PMF}(F)$. So it is reasonable to expect that there is $\phi$ with all angles between $\cal F^u\_\phi$ and $\cal F$ greater than some $\alpha>0.$ (But I don't have a rigorous argument for that.) Now by the work of Thurston (cf. the book above), the angles between $\phi^n(\gamma)$ and $\cal F^u\_\phi$ converge to $0$ as $n\to \infty.$ Consequently, $\phi^n(\gamma)$ is transverse to $\cal F$ for large $n$.


EDIT: To address the comment below, let us just assume that $\cal F$ admits an invariant measure of full support.



Can every curve be made transversal to a foliation by applying a pseudo-Anosov?
Given a full support measured foliation $\mathcal F$ and given a pseudo-Anosov $\phi$, what you want will work as long as $\mathcal F^u\_\phi$ can be isotoped to be transverse to $\mathcal F$.


The trouble is, this is not always possible. For general choice of $\mathcal F$ and $\phi$, there are a few things one can say. 


First, one can say that $\mathcal F$ and $\mathcal F^u\_\phi$ can be altered in their Whitehead equivalence classes so that they become transverse. 


Second, measured foliations themselves are annoying because of Whitehead equivalence, but you can do away with that annoyance in several ways. The most standard way is to work with measured geodesic laminations: the measured geodesic laminations associated to $\mathcal F$ and $\mathcal F^u\_\phi$ will always be transverse, unless they are equal. 


Another somewhat less satisfactory way is to work with the "partial measured foliations" that are used in one or two places in FLP. Every measured foliation $\mathcal F$ on $S$ is represented by a partial measured foliation $\mathcal F'$ which is canonical up to isotopy (slice apart all saddle connections, producing a partial measured foliation which has only annulus components and arational components; and then normalize the singularities on the arational components, with no interior saddle connections and with all boundary singularities being 3-pronged). Then one can say that $\mathcal F'$ and $\mathcal F^u\_\phi$ will be *almost* transverse after a homotopy of $\mathcal F^u\_\phi$; nonetheless there may still be violations of transversality at singularities, and notice that I wrote "homotopy" instead of "isotopy", because nearly parallel leaf segments of $\mathcal F^u\_\phi$ may be forced to coincide in order to achieve transversality with $\mathcal F$.


But to return to your actual question, given $\mathcal F$ does there exist $\phi$ so that $\mathcal F^u\_\phi$ can be isotoped to be transverse to $\mathcal F$? The answer is still no, because one can construct a measured foliation $\mathcal F$ which is not transversely recurrent: there is a simple closed curve $\sigma$ which is a concatenation of saddle connections of $\mathcal F$ such that there does not exist a simple closed curve transverse to $\mathcal F$ and having nonempty intersection with $\sigma$. (The reference for this construction, I think, is the Penner-Harer book "Combinatorics of Train Tracks").


Here's the construction of $\mathcal F$. Choose an essential, separating simple closed curve $c \subset S$, breaking $S$ into a union of two essential subsurfaces $S=S\_1 \cup S\_2$ with $c = S\_1 \cap S\_2$. On the surface $S\_i$, choose an arational measured foliation $\mathcal F\_i$ which is orientable and hence transversely orientable (we are assuming $S$ is oriented). We may, at first, assume that $\mathcal F\_i$ has canonical singularities: no interior saddle connections; all boundary singularities 3-pronged. But now alter $\mathcal F\_i$ as follows. Pick a boundary saddle connection $\alpha\_i$ for $\mathcal F\_i$, and alter $\mathcal F\_i$ by Whitehead equivalence, collapsing the arc $c - \alpha\_i$ to a point $p\_i$. Now the only boundary singularity of $\mathcal F\_i$ is $p\_i$. However, $p\_1$ and $p\_2$ might not be the same point, but if not then $c$ is a union of two saddle connections with endpoints $p\_1,p\_2$. Collapse one of those saddle connections to get the final example $\mathcal F$. 


Now one can prove that there is no simple closed curve $\gamma$ transverse to $\mathcal F$ that has nontrivial intersection with $c$. If $\gamma$ existed, write it as a concatenation
$$\gamma = \gamma\_1 \* ... \* \gamma\_{2K}
$$
where each $\gamma\_{2k+1}$ is contained in $S\_1$ transverse to $\mathcal F\_1$, and each $\gamma\_{2k}$ is contained in $S\_2$ transverse to $\mathcal F\_2$. The curve $\gamma\_1$ has endpoints on $c$ transverse to $\mathcal F$ at each endpoint. At the initial point the orientation on $\gamma\_1$ points inwards on $c = \partial S\_1$, and at the terminal point it points outward. However, that violates the fact that $\mathcal F\_1$ is transversely oriented.


