The answer is no. Here's an example.

Fix $Y$ and a closed point $p \in Y$. Blow up $p$. This is clearly smooth. Call the resulting scheme $X'$ and let $F$ denote the exceptional divisor of $f' : X' \to Y$. I am actually going to assume that $Y = \mathbb{A^3}$ in my mind.

Here's the idea then, within $X'$, choose a smooth curve $C$ which is tangent to $F$ at some point $p'$ in some "sufficiently funny way" and blow up the entire curve $C$.

Explicitly, if $X'$ has a chart $\text{Spec} k[x,y,z]$ and $F$ is given by the equation $z = 0$, then you can set $C = V(y, x^2 + z)$. This gives us $g : X \to X'$. Certainly $X$ is smooth and $f : X \xrightarrow{g} X' \xrightarrow{f'} Y$ is birational.

Now, I claim that the exceptional divisor of $f$ is *not* simple normal crossing. It is enough to show that the strict transform of $F$ in $X$ is not smooth. In our particular example, the strict transform of $F$ corresponds to the blowup of $(y,x^2)$ within $k[x,y]$. It is therefore easy to see that the strict transform of $F$ has a quadric cone singularity.