The following argument should also work when $X$ has klt singularities. 

By the result of Birkar-Cascini-Hacon-M$^{\rm c}$Kernan, we can find a birational model $f:X'\rightarrow X$ extracting only one divisor $E$ with $f(E)=W$. Since $-E$ is relative ample over $X$, there is an exact sequence,
$$f_*\mathcal{O}_{X'}=\mathcal{O}_X\rightarrow f_*\mathcal{O}_{E}\rightarrow R^1f_*\mathcal{O}_{X'}(-E)=0.$$ 
Since there is a factorization $$\mathcal{O}_X\rightarrow \mathcal{O}_W\rightarrow f_*\mathcal{O}_E,$$
the natural map $E\rightarrow W$ has connected fibers.
On the other hand, result of [Hacon-M$^{\rm c}$Kernan](https://projecteuclid.org/euclid.dmj/1178738561) asserts that $E\rightarrow W$ must have rationally connected fibers. 

The only situation that I know where one can find a birational section is  
when $\dim W=1$: These follows from the result of [Graber-Harris-Mazur-Starr](http://www.ams.org/leavingmsn?url=http://dx.doi.org/10.1090/S0894-0347-02-00402-2). To get a resolution of $X$, one can simply replace $X'$ by a higher model. 
  
However, if $W$ of $\dim W\geq2$, it seems to me the answer is related to weak approximation problem as handled in the last paper. As I remember, this is a nontrivial problem.   

When $W$ is also rationally connected, to have a rational section is still not an easy question. Artin-Mumford's conic bundle over $\mathbb{P}^2$ is a unirational but non-stably-rational variety.