I think the above answer is enough to prove $ht \; I \le ht \; I' +1 $ if $I \neq I'$. Pick a prime ideal $q$ that contains $I'$ such that $ht\; q = ht\; I'$. Recall that $q$ is homogenous since it is a minimal prime over a homogeneous ideal. By going modulo $q$ we may assume that $q = 0$. Now use the fact stated above by localizing at (0) as above. The image of $I$ is not zero since $I$ in $R_{(0)}$ is not in $q$. And it is not the whole ring since we are only inverting homogeneous elements that are not in $q$ which contains $I'$. Then since $R_{(0)}$ is a PID in particular one diemsimonal domain, the height of the image of $I$ is 1. Now apply the above result again to conclude that $ht I \le ht I' + 1$.
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I don't think $ht \;I = ht\; I' + 1$ if $I \neq I'$ in general. Here is an example. Let $R = k[x_1,\dots,x_n]_{(x_1,\dots,x_n)}$ and $m = (x_1,\dots,x_n)R$. Let $I = (f)+m^s$ where $s$ is large and $f \in m$ is an non-homogeneous polynomial whose terms have degree less than $s$. Then we have $m^s \subset I' \subset I \subset m$. Hence $ht \; I' = ht \; I \; (= d)$.
I needed to say "if $ht\; I \neq ht \;I'$" instead of "if $I \neq I'$" in my original answer. This happens because there may exist a homogeneous prime ideal that is minimal over both $I$ and $I'$. In other words, the converse of the statement "every minimal prime of a homogeneous ideal is again homogeneous" is not true.