Why lower bounds for boolean Circuits does not imply arithmetic circuits lower bounds
For arithmetic circuits over $\mathbb{Z}$ your argument is exactly right. The same argument works for arithmetic circuits over $\mathbb{Q}$ which don't use any fractions $a/b$ where $b$ is even.

However, the argument no longer works if you talk about arithmetic circuits over other rings, such as: general arithmetic circuits over $\mathbb{Q}$ (i.e. without the restriction above), $\mathbb{R}$, algebraic number fields, $\mathbb{C}$, or finite fields $\mathbb{F}_{q}$ with $q \neq 2$. 

(This is essentially the same reason that in algebraic geometry $\mathbb{Z}$ is often considered of so-called "mixed characteristic," rather than characteristic zero.)

However, depth 3 Boolean lower bounds for circuits with {AND,OR,NOT} are less easily related to lower bounds for arithmetic circuits over $\mathbb{Z}$. (Yes, {AND,XOR} is a complete basis, but typically depth 3 circuits over {AND, OR, NOT} you consider NOT gates free, whereas implementing NOT with XOR you're then using an XOR gate, which you actually count. Similarly, although $a \vee b = \neg (\neg a \wedge \neg b)$, when you implement this single OR gate with AND and XOR, you get a little gadget of depth 3.)

The general statement is: let $f$ be a polynomial with coefficients in a ring $R$, and suppose $\varphi\colon R \to S$ is a ring homomorphism. By applying $\varphi$ to every coefficient of $f$ you get a polynomial with coefficients in $S$, which I'll denote $f_S$. Then a lower bound for computing $f_S$ by $S$-arithmetic circuits implies the same lower bound for computing $f$ by $R$-arithmetic circuits.