**Correction.** I just realized that there are examples where the geometric generic fiber is **NOT** generically reduced. In all of my comments and the answer below, I was assuming that the geometric generic fiber is generically reduced. When that is true, then the geometric generic fiber is integral. However, without the hypothesis (or some other hypothesis that implies this hypothesis), the argument below only proves that the geometric generic fiber is an irreducible, LCI scheme. It may be everywhere nonreduced. There is a numerical condition that will guarantee this. For any ample divisor class $A$ on $X$, and for any ample divisor class $B$ on $Y$, if the intersection number $A^{\text{dim}(X)-\text{dim}(Y)}.(f^*B)^{\text{dim}(Y)}$ is prime to $p$, then the geometric generic fiber of $f$ is generically reduced. I noticed the counterexamples below by looking for examples where $p$ divides this integer.

**Original post.**
For a normal, integral scheme $X$, for an integral scheme $Y$, for a proper, locally finitely presented morphism $f: X\to Y$, if the natural map $f^\# : \mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism, then the geometric generic fiber of $f$ is integral and LCI (thus Cohen-Macaulay), but quite possibly not $R1$, i.e., not normal.

Denote by $K(Y)$ the function field of $Y$. Denote by $X_\eta$ the fiber product of $f$ and the natural morphism $i:\text{Spec}\ K(Y) \to Y$. Since $X$ is a normal scheme and $X_\eta$ is obtained by localization, also $X_\eta$ is a normal $K(Y)$-scheme. Also the natural map $K(Y)\to H^0(X_\eta,\mathcal{O}_{X_\eta})$ is an isomorphism since it is the generic fiber of the isomorphism $f^\#$.

Denote by $K(Y)^{\text{sep}}/K(Y)$ the separable closure of $K(Y)$. Denote by $X_{\eta^s}$ the base change of $X_\eta$ to $K(Y)^{\text{sep}}$. Since $K(Y)^{\text{sep}}/K(Y)$ is a limit of etale extensions, the same holds for $X_{\eta^s}\to X_\eta$. In particular, $X_{\eta^s}$ is a normal scheme. Since the base change is flat, also $K(Y)^{\text{sep}} \to H^0(X_{\eta^s},\mathcal{O}_{X_{\eta^s}})$ is still an isomorphism. Thus, $X_{\eta^s}$ is still a normal, integral $K(Y)^{\text{sep}}$-scheme.

Finally, the field extension to the algebraic closure $\overline{K(Y)}/K(Y)^{\text{sep}}$ is purely inseparable. Thus for the corresponding base change $X_{\overline{\eta}}$ of $X_{\eta^s}$, the morphism $X_{\overline{\eta}} \to X_{\eta^s}$ is also purely inseparable. In particular, it is a homeomorphism (with respect to the Zariski topologies). Since $X_{\eta^s}$ is irreducible, also $X_{\overline{\eta}}$ is irreducible. As explained in my comments above, $X_{\overline{\eta}}$ is LCI and generically reduced, hence (everywhere) reduced. Thus, the geometric generic fiber $X_{\overline{\eta}}$ is integral and LCI. However, there are examples (like quasi-elliptic fibrations) where $X_{\overline{\eta}}$ is not $R1$.

**Edit.** The OP did not ask this, but I (and my advisees) have occasionally found it useful. In the setting above, since the divisorial part of the closed subscheme $X_{\eta}^\text{sing} \subset X_{\eta}$ (cut out by the Fitting ideal of the sheaf of relative differentials, for example) must be inseparable over $K(Y)$, this forces certain numerical invariants to be divisible by the characteristic $p$. So if you study varieties with those numerical invariants equal to a specified (nonzero) integer, if you exclude "small characteristics" that divide that integer, then $X_{\overline{\eta}}$ will be normal.

**Counterexample to Generic Reducedness.** Let $k$ be a field of characteristic $p$. Let $Y$ be $\mathbb{P}^2_k = \text{Proj}\ k[s_0,s_1,s_2]$. Let $T$ be $\mathbb{P}^2_k = \text{Proj}\ k[t_0,t_1,t_2]$. Let $X\subset T\times_{\text{Spec}\ k} Y$ be the closed subscheme $\text{Zero}(t_0^ps_0 + t_1^ps_1 + t_2^ps_2)$. The projection $\text{pr}_T:X\to T$ is a smooth, projective morphism; it is Zariski locally a $\mathbb{P}^1$-bundle. Define $f:X\to Y$ to be the other projection. This morphism is nowhere smooth. The geometric generic fiber is everywhere nonreduced.

The argument above does prove that the geometric generic fiber is irreducible. Moreover, if the geometric generic fiber is generically reduced, then it is everywhere reduced. However, there are examples where the geometric generic fiber is everywhere nonreduced.