I don't think this is true. A simple example is that the self-intersection of a divisor changes when you take its strict transform under a flop. In more detail:

Suppose that $\phi : X \dashrightarrow Y$ is a standard flop between smooth threefolds. Here's what I mean: there is a rational curve $C \subset X$ with normal bundle $\mathcal O(-1) \oplus \mathcal O(-1)$, we blow it up to get $f : Z \to X$ with an exceptional divisor $E$ isomorphic to $\mathbb P^1 \times \mathbb P^1$, which can be blown down along the other ruling via $g : Z \to Y$, contracting $E$ to a curve $C^+$. The induced map $\phi : X \dashrightarrow Y$ is birational, with no exceptional divisors in either direction.

Take $H$ very ample on $Y$, with $D$ its pullback via $\phi$ to $X$ (equivalently, its strict transform). This $D$ has base locus along the flopping curve, and its linear system gives the flop we want.

Now, $f^\ast D = g^\ast H + aE$, where $a = H \cdot C^+$ (which is greater than $0$, since $H$ is ample). Take the top self-intersection of both sides. On the left, we just have $D^3$. On the right, we have $H^3 + 3 a \, (g^\ast H)^2 E + 3 a^2 \, (g^\ast H) E^2 + a^3 E^3 = H^3 -3a^2 \, (H \cdot C^+) + 2a^3 = H^3 - a^3$. So $D^3 \neq H^3$.