Your question is contained in Theorem 1.5.8 of Bruns/Herzog "Cohen-Macaulay-Rings". The basic reason is that for any non-graded prime $p$ one has $\text{ht}(p/p^*) = 1$. To see this replace $R$ by $R/p^*$ and assume that $p^*=0$ and that $p$ does not contain a nonzero homogeneous element. Now invert all homogeneous elements, passing to $R_{(0)}$ and use the fact that in a graded ring $R$ all homogeneous elements are invertible if and only if $R$ is a field or $R_0 = k$ is a field and $R=k[t,t^{-1}]$ for some homogeneous element $t\in R$ of positive degree and transcendental over $k$ (BH, Lemma 1.5.7). Since $R_{(0)}$ has the non-zero prime $pR_{(0)}$ we get $\text{ht}(p) = 1$.

The answer follows from Theorem 1.5.8 of Bruns/Herzog "Cohen-Macaulay-Rings": The basic reason is that for any non-graded prime $p$ one has $\text{ht}(p/p') = 1$. To see this replace $R$ by $R/p'$ and assume that $p'=0$ and that $p$ does not contain a nonzero homogeneous element. Now invert all homogeneous elements, passing to $R_{(0)}$ and use the fact that in a graded ring $R$ all homogeneous elements are invertible if and only if $R$ is a field or $R_0 = k$ is a field and $R=k[t,t^{-1}]$ for some homogeneous element $t\in R$ of positive degree and transcendental over $k$ (BH, Lemma 1.5.7). Since $R_{(0)}$ has the non-zero prime $pR_{(0)}$ we get $\text{ht}(p) = 1$. That all said, choose $p'\supset I'$ with $\text{ht}(p') = \text{ht}(I')$. The first inequality follows since $\text{ht}(IR_{(0)})=1$ and in the original ring we can conclude $\text{ht}(I) = \text{ht}(p') + 1 = \text{ht}(I') + 1$. The second inequality is obvious since $I'\subset I$.