As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper.  

For the first question then:  sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$).  This condition is basically that the non-singular $X$ has *rational singularities*.  In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not).  Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic [http://front.math.ucdavis.edu/0911.3599][1].

For the second question *"fibres are not unions of projective spaces"*, you've already seen one non-normal example.  I believe that fibers of such maps should always have unirational irreducible components (unirational means birational to an image of $\mathbb{A}^n$ under some map) but nothing more.  I'm sure one can cook up normal examples too, but I don't know of any with the possible exception of the following counter-example to your third question (which I haven't checked).

The last question, *"is it always true that $H^n(Y,O_Y)=0$?"*, here's an example which this time is normal.  It is Section III of Cutkosky's [*A new characterization of rational surface singularities*][2]
Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in.

**Note:**  In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this [answer on math overflow][3].


  [1]: http://front.math.ucdavis.edu/0911.3599
  [2]: http://link.springer.com/article/10.1007/BF01233425
  [3]: http://mathoverflow.net/a/43415/3521