How much can small modifications change the nef cone?
In the world of Calabi-Yau (as opposed to Fano) varieties, one does not expect miracles of Mori dream space type. A specific example which gives a positive answer to your first question is described in the paper http://xxx.lanl.gov/abs/math/0102055 of Michael Fryers: a Calabi-Yau threefold (a degenerate quintic) which has some small resolutions having finite polyhedral nef cone, and some having an infinitely generated cone, which is locally rational polyhedral away from a single point on the boundary. The geometry, related to the Horrocks-Mumford bundle and abelian surfaces in projective four-space, is very beautiful.
Let me start by a disclaimer: I misread the question and wrote what's below thinking that it was relevant. Then, just before posting it I looked at the question one more time and realized the mistake.
As it happens, the statement and proof below is mathematically correct (I think), except that it does not answer the question, or one could perhaps even say that it does not have much to do with the question and probably everyone before me who have looked at this question knows what I wrote down here.
At the same time, I would hate to just erase what I typed up and perhaps someone finds this useful, so I will just post it. If some MO users think this is inappropriate, I will be happy to erase it, just let me know.
So, here come some mathematical ramblings loosely related to the question above.
(The point is that there does not exist a morphism, but the question was about rational maps).
There does not exist a morphism that is an SQM. This is because if $f:X\to Y$ is a small morphism, then $Y$ is not $\mathbb Q$-factorial. In fact a little more seems to be true:
Proposition. Let $f:X\to Y$ be a proper birational morphism and assume that $X$ is quasi-projective and $Y$ is $\mathbb Q$-factorial. Then the exceptional set of $f$ is of pure codimension $1$.
Proof. Let $H\subset X$ be a very ample Cartier divisor on $X$ and consider $f_*H$ the cycle theoretic image of $H$. This is a priori only a Weil divisor, but since $Y$ is $\mathbb Q$-factorial, it is $\mathbb Q$-factorial and replacing $H$ with an appropriate multiple we may assume that it is in fact Cartier.
Now consider $F:=f^*f_*H-H$ and notice that a) $F$ is effective b) $-F$ is $f$-ample. These show that $F$ must be exceptional. This already proves that $f$ cannot be small, but we can prove the stronger statement stated above.
${\rm Supp}\, F$ is obviously of pure codimension $1$, so if we prove that the exceptional set of $f$ is equal to this, then we are done. We have already seen that ${\rm Supp}\, F$ is contained in the exceptional set, so we only need to demonstrate that $f$ is an isomorphism on the complement of $F$.
Now, observe that by the definition of $F$, we have that $$ \mathcal O_X(-F)\simeq \mathcal O_X(H)\otimes\mathcal O_X(-f^*f_*H) $$ and $$ \begin{multline*} f^*f_* \mathcal O_X(-F)\simeq f^*f_*\mathcal O_X(H-f^*f_*H)\simeq \\ \simeq f^*\left(f_*\mathcal O_X(H)\otimes \mathcal O_Y(-f_*H)\right)\simeq f^*f_*\mathcal O_X(H)\otimes \mathcal O_Y(-f^*f_*H) \end{multline*} $$
Furthermore, since $H$ is very ample, it is generated by global sections, so $$ f^*f_*\mathcal O_X(H) \to \mathcal O_X(H) $$ is surjective. Then it follows from the above displayed lines that $$ f^*f_*\mathcal O_X(-F) \to \mathcal O_X(-F) $$ is also surjective. This implies that $f^{-1}(f({\rm Supp}\, F)) = {\rm Supp}\, F$. It follows then that $H$ and $f^*f_*H$ agree on $X\setminus {\rm Supp}\, F$. In particular, $f^*({\rm some divisor})$ is ample on this $X\setminus {\rm Supp}\, F$. Finally this implies that then $f$ has to be an isomorphism on $X\setminus {\rm Supp}\, F$. Q.E.D.