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The assertion is false. For $n\ge 2$, take the affine hypersurfaceIt is an integral scheme, and its only possible singular point is the origin, which has codimension two. Hence it is regular in codimension one; as a hypersurface it satisfies Serre's condition $(S_2)$, so the Serre criterion for normality also makes it a normal scheme. The Divisor class group of an A-type surface singularity isgenerated by $V(x,z)$. Thus a closed affine subscheme satisfying $(\star )$ can have nonzero torsion in its class group.

For a point $p\in {\mathbb{A}}_k^1$, let $a$ be the image of $t$ in its residue field $\kappa (p)$. The scheme-theoretic fiber isIf $p\ne (t)$, then $a\ne 0$. The quadratic polynomial $xy-a{z}^2$ is irreducible. Indeed, after setting $z=0$, any hypothetical linear factors must restrict, up to nonzero scalars, to $x$ and $y$; comparing the $xz$ and $yz$ coefficients then forces both $z$ coefficients to vanish, contradicting the nonzero $z^2$ coefficient. Hence its homogeneous coordinate ring is an integral domain, so $Y_p$ is an integral scheme.

At the origin $p=(t)$, the fiber is $V_{+}(xy)$, the union of the two distinct projective lines $x=0$ and $y=0$, and is therefore not irreducible. It is nevertheless a reduced scheme because the ideal $(xy)=(x)\cap (y)$ equals its radical. Every other fiber is integral and hence reduced. Thus the fiber is integral exactly away from the origin, and it is reduced at every point of ${\mathbb{A}}_k^1$.