Regarding the suggestion in my comment, there is a subtle point regarding my proposed pencil of curves; the Euler identity fails, so one has to check by hand that the singular members occur only for $t=0$ and $t=\infty$. In fact, this is not always the case. However, I checked this in one case.

Let $k$ be an algebraically closed field of characteristic $2$. Let $a,b\in k$ be elements such that none of $a,b,a+b$ equals $0$. Denote $\mathbb{P}^1_k$ by $\overline{Y}$, and let $[s,t]$ be homogeneous coordinates on $\mathbb{P}^1_k$. Let $[u,v,w]$ be homogeneous coordinates on $\mathbb{P}^2_k$. Let $\overline{X}$ be the Cartier divisor in $\mathbb{P}^1_k \times_k \mathbb{P}^2_k$ with defining equation $f(s,t;u,v,w) = suvw(u+v+w) - t(au+bv-(a+b)w)^4$. Consider the projection morphism $\overline{\pi}:\overline{X}\to \overline{Y}$. By direct computation, the only singular points of the morphism occur where $([s,t],[u,v,w])$ equals one of
$$
([1,0],[1,0,0]), ([1,0],[0,1,0]), ([1,0],[0,0,1]),
$$
$$
([1,0],[1,1,0]), ([1,0],[1,0,1]), ([1,0],[0,1,1]), ([0,1],[u,v,w]).
$$
The tricky coordinates, of course, are $[u,v,w]=[1,1,1]$. However, the conditions on $a$ and $b$ guarantee this is contained in the member $[s,t]=[0,1]$ of the pencil. So this is a pencil to which my comments above apply. Define $Y\subset \overline{Y}$ to be the open complement of $\{[1,0],[0,1]\}$. Define $X\subset \overline{X}$ to be the inverse image of $Y$ under $\overline{\pi}$. Finally, define $\pi:X\to Y$ to be the restriction of $\overline{\pi}$. Then $\pi$ is a proper, smooth morphism.

To summarize the comments, by Deligne-Fulton the tame fundamental group of the open complement $U:= \mathbb{P}^2 \setminus Z(uvw(u+v+w)(au+bv-(a+b)w))$ is Abelian. This is also a dense open subset of $X$. Thus the induced map from the étale fundamental group of $U$ to the étale fundamental group of $X$ is surjective. In particular, the tame fundamental group of $X$ is also Abelian. The geometric generic fiber $X_{\overline{\eta}}$ of $\pi$ is a smooth, genus $3$ curve. Therefore its tame fundamental group is non-Abelian. So the kernel of the homomorphism of ~~étale~~ tame fundamental groups $\pi_1^t(X_{\overline{\eta}})\to \pi_1^t(X)$ is non-Abelian.

$\textbf{Edit}.$ I am a little worried now about the distinction between the tame fundamental group and the full étale fundamental group in this argument. Take any finite group $G$ of order prime to $p$. There is an injective homomorphism into a finite symmetric group $S_{p^r}$. Yet the maximal prime-to-$p$ quotient of $S_{p^r}$ is just $S_2$ (or trivial if $p$ equals $2$). So perhaps the kernel of the map of tame fundamental groups is much larger than the kernel of the map of full étale fundamental groups.

$\textbf{Second Edit}.$ I looked in the article of Friedlander that Will Sawin linked. Friedlander also needs to work with completions of the higher étale homotopy groups at primes different from the characteristic; in particular, this factors through the maximal prime-to-$p$ quotient. My example above also works if we look at the maximal prime-to-$p$ quotients of the étale fundamental group, or if we look at an $\ell$-completion for $\ell$ a prime different from $p$, since both of these factor through the tame fundamental group. So I feel that this example does explain why Friedlander did not state his theorem for $\pi_1$ as well as for $\pi_n, n\geq 2$.