Base change of a finite morphism Let $X$, $Y$, $Z$ be integral schemes of finite type over a field $K$ (i.e., locally affine opens are finitely-generated algebras over $K$). Suppose we have the following condition$\colon$
$f \colon Y \to X$ is finite.
We consider a dominant morphism $g \colon Z \to X$. 
Q. Suppose that ${\cal O}_Y \cap {\cal O}_Z = {\cal O}_X$. Then is it possble that $Y \times_X Z$ is not integral while $Y \times_X \eta_Z$ being integral ?
As in the answer below in the case $Y = Z$, there is an example as $Y$ being a normalisation of $X$.
 A: Edit. I edited the post below to address the OP's (repeated) modifications of the question.
Answer to original question without any hypothesis that $\mathcal{O}_X$ equals $\mathcal{O}_Y\cap \mathcal{O}_Z.$
I am just posting my comment as an answer.  Let $X$ be an integral scheme that is not normal, e.g., $\text{Spec}\ k[u,v]/\langle u^2 - v^3\rangle$ for a field $k$ of characteristic $\neq 2,3$.  By a corollary of the Noether normalization theorem, the normalization morphism $\nu:X^{\text{nor}}\to X$ is a finite morphism.  Denote by $U\subset X$ the maximal open subscheme over which $\nu$ is flat, and denote by $C\subset X$ the closed complement of $X$.  By openness of the normal locus (should be in Section 24 of Matsumura, "Commutative Ring Theory"), since the generic point $\text{Spec}\ k(X)$ is normal, the open $U$ is dense.  Over $U$, the morphism $\nu$ is an isomorphism.  For every point $p\in C$, the fiber of $\nu$ over $p$ has length $\ell(p)\geq 2$.
Let both $f:Y\to X$ and $g:Z\to Y$ equal $\nu$.  Then $X^{\text{nor}}\times_X X^{\text{nor}}\to X$ is a finite morphism that is an isomorphism over $U$.  Since $\nu$ is finite, hence affine, it is separated.  Thus, the diagonal morphism, $$\Delta:X^{\text{nor}}\to X^{\text{nor}}\times_X X^{\text{nor}},$$ is a closed immersion.  In particular, when we restrict over the open $U$, this closed immersion is an isomorphism.  Hence, this closed immersion is one irreducible component of the reduced scheme of $X^{\text{nor}}\times_X X^{\text{nor}}$, and it is the unique irreducible component that intersects the inverse image of $U$.
However, the length of the fiber of $\Delta(X^{\text{nor}})$ over $p\in C$ equals $\ell(p)$, whereas the fiber of $X^{\text{nor}}\times_X X^{\text{nor}}$ equals $\ell(p)^2 > \ell(p)$.  Thus, the closed immersion $\Delta$ fails to be an isomorphism over every point of $C$.  So $X^{\text{nor}}\times_X X^{\text{nor}}$ is not integral.  
In the example $X=\text{Spec}\ k[u,v]\langle u^2 - v^3\rangle$, the scheme $X^{\text{nor}}\times_X X^{\text{nor}}$ is irreducible but not reduced.  In the example $X=\text{Spec}\ k[u,v]/\langle u^2 - v^2(v-1) \rangle$, the scheme $X^{\text{nor}}\times_X X^{\text{nor}}$ is reducible (the extra irreducible components come from pairs of distinct points of $X^{\text{nor}}$ that are "separated apart" from branches of singularities of $X$).
Answer to modified question with hypothesis that $\mathcal{O}_X$ equals $\mathcal{O}_Y\cap \mathcal{O}_Z.$  All of the following rings will be algebras over the power series ring $k[[t]]$.  The schemes below are reduced and equidimensional, but not integral.  It is straightforward to find integral schemes that are finite type over $k$ and such that the schemes below are obtained via base change by $\text{Spec}\ \widehat{\mathcal{O}}_{X,p} \to X.$  For instance, this works for the plane curve $X=\text{Spec}\ k[u,v]/\langle v^3-(u^3+u^4)\rangle$ and the "triple point" $p=\langle u,v\rangle.$
Let $A$ be $k[[t]]\times k[[t]]\times k[[t]].$  Denote by $e_1,$ $e_2,$ and $e_3$ the elements $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.  Thus, $A$ is the finite $k[[t]]$-algebra, $$A=k[[t]][e_1,e_2,e_3]/\langle e_1^2-e_1,e_2^2-e_2,e^3_2-e_3, e_1+e_2+e_3-1, e_1e_2,e_1e_3,e_2e_3\rangle.$$    
Let $\mathcal{O}_X,$ $\mathcal{O}_Y,$ and $\mathcal{O}_Z$ be the following $k$-subalgebras of $A,$ $$\mathcal{O}_X := k[[t]][te_1,te_2,te_3], \ \ \mathcal{O}_Y := k[[t]][e_1,te_2,te_3], \ \ \mathcal{O}_Z := k[[t]][te_1,te_2,e_3].$$  Thus, an element of $A$, $$a = (c_1 ,c_3,c_3) + tb, \ \ (c_1,c_2,c_3)\in k\times k \times k,\ \ b\in A,$$ is in $\mathcal{O}_X$, resp. $\mathcal{O}_Y$, resp. $\mathcal{O}_Z$, if and only if $c_1=c_2=c_3$, resp. $c_2=c_3$, resp. $c_1=c_2$.  The intersection of $\mathcal{O}_Y$ and $\mathcal{O}_Z$ in $A$ equals $\mathcal{O}_X$, i.e., if both $c_2$ equals $c_3$ and also $c_1$ equals $c_2$, then $c_1=c_2=c_3$.  Also, $\mathcal{O}_Y$ and $\mathcal{O}_Z$ generate $A$: the element $e_1$ is in $\mathcal{O}_Y$, the element $e_3$ is in $\mathcal{O}_Z$, and thus the element $e_2=1-(e_1+e_3)$ is in the subring that they generate.
As extensions of $\mathcal{O}_X$, the rings $\mathcal{O}_Y$, resp. $\mathcal{O}_Z$, equal, $$\mathcal{O}_Y \cong \mathcal{O}_X[f_1]/\langle f_1^2-f_1,tf_1-te_1\rangle,\ \  \mathcal{O}_Z \cong \mathcal{O}_X[f_3]/\langle f_3^2-f_3,tf_3-te_3\rangle.$$
For the maximal ideal $\mathfrak{m}_X = \langle te_1,te_2,te_3 \rangle$ in $\mathcal{O}_X$ with residue field $k$, both of the fiber rings $\mathcal{O}_Y/\mathfrak{m}_X\cdot \mathcal{O}_Y$ and $\mathcal{O}_Z/\mathfrak{m}_X\cdot \mathcal{O}_Z$ have dimension $2$ as $k$-vector spaces, generated by $1$ and $f_1,$ resp. generated by $1$ and $f_3.$  Thus, the fiber ring $(\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z)/\mathfrak{m}_X\cdot (\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z)$ has dimension $4$ as a $k$-vector space, generated by $$ 1\otimes 1, \ \ 1\otimes f_3, \ \ f_1\otimes 1, \ \ f_1\otimes f_3.$$  On the other hand, $A/\mathcal{m}_X\cdot A$ has dimension $3$ as a $k$-vector space generated by $$1,\mathcal{e}_1, \mathcal{e}_3.$$   Thus, the kernel of the surjective $k[[t]]$-algebra homomorphism, $$\mathcal{O}_Y\otimes_{\mathcal{O}_X} \mathcal{O}_Z\to A,$$ is an ideal of length $1$.  Therefore $\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z$ is not an integral domain. 
Explicitly, the kernel is generated by the element $f=f_1\otimes f_3.$  Note that this is an idempotent element, so $\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z$ has an extra irreducible component that does not dominate any irreducible component of $\mathcal{O}_X$.  In fact $f$ is annihilated by $t$ since $$t\cdot f = (tf_1)\otimes f_3 = (1\cdot te_1)\otimes f_3 = 1 \otimes (te_1\cdot f_3) = 1\otimes 0 = 0.$$
A: The assumption $\mathcal O_Y\cap\mathcal O_Z=\mathcal O_X$ needs to be made more precise. In order for the intersection $\mathcal O_Y\cap\mathcal O_Z$ of two rings to make sense, they must both be contained in some larger set. This larger set must be specified here. One option is to consider $\mathcal O_Y$ and $\mathcal O_Z$ as subrings of $\mathcal O_Y\otimes_{\mathcal O_X}\mathcal O_Z$. But then the condition $\mathcal O_Y\cap\mathcal O_Z=\mathcal O_X$ is satisfied, so the above example of Jason Starr still applies.
If $X$, $Y$ and $Z$ happen to be integral and the maps $f$ and $g$ happen to be birational, as in Jason Starr's example, then we can take the intersection $\mathcal O_Y\cap\mathcal O_Z$ inside their common field of fractions. This more restricted setting rules out Jason Starr's example.
