The first question has been answered by Michael Renardy, and the answer is no. The second answer should have a positive answer. To avoid any domain issues, let me explain rather why $A^2 \leq B^2$ implies that $A B^{-1}$ has norm $\leq 1$ (for bounded operators $B^{-1} A$ is the adjoint of $A B^{-1}$ so it also has norm $\leq 1$, but I never work with unbounded operators so not sure whether this is true always).

For $\xi$ in the domain of $B$, if $\eta = B \xi$,
$$\|A B^{-1} \eta \|^2 = \| A \xi\|^2 = \langle A^2 \xi,\xi\rangle \leq \langle B^2 \xi,\xi\rangle = \|\eta\|^2.$$
This shows that $A B^{-1}$ extends to a norm $\leq 1$ operator on the closure of the image of $B$, which (I guess since you write $B^{-1}$) is assumed to be the whole space.

Another way to write the same proof is, for $A,B \geq 0$,
$$ A^2 \leq B^2 \iff B^{-1} A B^{-1} \leq 1 \iff \|A B^{-1}\| \leq 1.$$
The first equivalence is just the fact that the operation $ X \mapsto B X B$ preserves positive operators.