As Kevin says, you can use Neron-Ogg-Shafarevich. Here's an alternative proof.
We can assume that the local field $K$ is algebraically closed and complete for the valuation $v$. Then $E$ is $v$-adically analytically isomorphic to a Tate curve, which means that there is a $v$-adic analytic isomorphism $E(K)\to K^*/q^{\mathbb{Z}}$, where $q\in K^*$ satisfies $|q|_v<1$. Further, $|j(E)|_v = |q|_v^{-1}$. Let $E'$ be isogenous to $E$, say $\phi:E\to E'$. In the analytic model, the kernel of $\phi$ is generated by two elements $\zeta$ and $Q$, where $\zeta$ is an $n$'th root of unity and $Q$ is some root of $q$, say $Q^m=q$. Then$$E'(K) \cong K^*/\zeta^{\mathbb{Z}}Q^{\mathbb{Z}} \cong K^*/Q^{n\mathbb{Z}},$$ where the second isomorphism is raising to the $n$'th power. Thus $E'$ has a Tate parametrization, and the analytic formula gives $$|j(E')|_v = |Q^n|_v^{-1} = |q|^{-n/m}_v > 1.$$ Hence $E'$ has non-integral $j$-invariant.
You can find a discussion of Neron-Ogg-Shafarevich in my Arithmetic of Elliptic Curves, Chapter VII, Section 7, but there's one step (the hardest step) of the proof that is not done there. That step is in my Advanced Topics in the Arithmetic of Elliptic Curves, as is a detailed discussion of Tate models in Chapter V.
