I am posting as an "answer" the generalization of the example in my comment above. The ring $R$ has your property if the ring $R$ has the following property: for every minimal prime ideal $\mathfrak{p}$, there exists $x_{\mathfrak{p}}\in R\setminus \mathfrak{p}$ that annihilates $\mathfrak{p}$. In this case, let $I'$, resp. $J$', be the ideal generated by all those elements $x_{\mathfrak{p}}$ for minimal primes $\mathfrak{p}$ that do not contain $I$, resp. $J$.
By hypothesis, no minimal prime $\mathfrak{p}$ contains both $I$ and $J$. Thus, for every generator $x_{\mathfrak{p}}$ of $I'$ and for every general $x_{\mathfrak{q}}$ of $J'$, $\mathfrak{p}$ does not equal $\mathfrak{q}$. Thus, there exists $y_{\mathfrak{p}}\in \mathfrak{p}\setminus \mathfrak{p}\cap \mathfrak{q}$. Since $x_{\mathfrak{p}}y_{\mathfrak{p}} = 0$, and since $y_{\mathfrak{p}}$ is not in $\mathfrak{q}$, necessarily $x_\mathfrak{p}$ is in $\mathfrak{q}$. Thus, since $x_{\mathfrak{q}}$ annihilates $\mathfrak{q}$, $x_{\mathfrak{p}}\cdot x_{\mathfrak{q}}$ equals $0$. Therefore $I'J'$ equals $\{0\}$. Yet for every minimal prime $\mathfrak{p}$, if $\mathfrak{p}$ contains $I$, resp. $J$, then $\mathfrak{p}$ does not contain $I'$, resp. $J'$. Therefore $I+I'$ and $J+J'$ are contained in no minimal primes.
Please note, for every ring $R$, for every minimal prime $\mathfrak{p}$, for every minimal prime $\mathfrak{q}\neq \mathfrak{p}$, there does exist $y_{\mathfrak{q}}\in \mathfrak{q}\setminus \mathfrak{p}$. If $R$ is Noetherian, then there are only finitely many such minimal primes $\mathfrak{q}$ different from $\mathfrak{p}$, and the product $x_{\mathfrak{p}}$ of the finitely many elements $y_{\mathfrak{q}}$ is an element that is not in $\mathfrak{p}$, yet it is in every other minimal prime. Thus $x_{\mathfrak{p}}\cdot \mathfrak{p}$ is contained in every minimal prime. Assuming that the ring $R$ is also reduced, $x_{\mathfrak{p}}\cdot \mathfrak{p}$ equals $\{0\}$. Thus the hypothesis above holds for reduced, Noetherian rings.