Looking for an example of non coherent sheaf $F$ over $X\subset P_R^n$ s.t. $H^{n+1}(X,F)\neq 0$. This is related to Ueno's Algebraic Geometry, Thm 6.21.
Thm 6.21 Let $S$ be a graded ring s.t. $S$ is f.g. over $R$ as an algebra and its finite generators are in degree 1. Assume $R$ is noetherian. Then for a coherent sheaf $F$ over $X=Proj(S)$, $H^p(X,F)$ is finite $R-$module. Furthermore, there is $m_0\geq 0$ s.t. $\forall m\geq m_0$, $H^p(X,F(m))=0$. For any coherent sheaf $F$ over $X$, there is $n_0$ s.t. $\forall p\geq n_0$ $H^p(X,F)=0$. 
$\textbf{Q:}$ Since the proof uses coherency of $F$ via resolution of $F$ by $O_X$'s, I am looking for an example of $F$ that is not coherent s.t $H^i(X,F)$ is not finite over $R$ for some $i$ and $H^p(X,F)\neq 0$ when $p=n+1$. 
$\textbf{Q':}$ How bad is situation if $F$ is quasicoherent not coherent?
$\textbf{Q'':}$ If $F$ is not coherent, how bad do I fail $H^p(X,F(m))=0$ for all $p\geq 0$ when $m$ large. 
 A: Q: For an example of non coherent sheaf such that $H^{n+1}(X,F)\neq 0$. You can take $n=0$ and $X=\mathbb{P}^0_R$. So we are looking for a sheaf $F$ on $\operatorname{Spec}R$ such that $H^1(\operatorname{Spec}R,F)\neq 0$. This is easy to produce : take $R=\mathbb{C}[X]$. In fact, $\operatorname{Spec}R$ as a topological space is isomorphic to $\mathbb{P}^1_{\mathbb{C}}$ (both are an uncountable set with the cofinite topology and a generic point). Since there exist a sheaf on $\mathbb{P}^1_{\mathbb{C}}$ with non-vanishing $H^1$, you can transport it (using the homeomorphism) to a sheaf with non-vanishing $H^1$ on $\operatorname{Spec}R$.
For an example of non coherent sheaf such that $H^i(X,F)$ is not a finite-type $R$-module, just take an infinite direct sum of $\mathcal{O}_X$ (it has a non-finite $H^0$).
Q' : Coherence vs quasi-coherence is mostly a "finiteness" condition. The quasi-coherence gives the vanishing result, the coherence gives the fact that $H^i$ is of finite type. The coherence also gives that there exists $m_0$ such that for all $m\geq m_0$, $F(m_0)$ has no higher cohomology.
Q'' : Take $G$ a sheaf with non vanishing $H^i$ and $F=\bigoplus_{n\in\mathbb{Z}} G(n)$. Then $F$ has a non vanishing $H^i$ and for all $m\in\mathbb{Z}$, $F\simeq F(m)$ so that $F(m)$ also has a non-vanishing $H^i$.
A: There's some disagreement between your title question and your body questions. I'll address the title question first.

Theorem (3.6.5 of Grothendieck, Sur quelques points d'algèbre homologique, Tohoku Mathematical Journal vol 9 pg 119-221, 1957):
Let $X$ be a noetherian topological space. If $\dim(X) \leq d$, then $H^p(X,\mathcal{F})=0$ for all $p>d$ and any abelian sheaf $\mathcal{F}$ on $X$.

If $R$ is zero dimensional as a ring, then your $X$ is noetherian of dimension $\leq n$, we have that $H^{n+1}(X,\mathcal{F})=0$ for any $\mathcal{F}$ by the theorem. So you won't be able to find the sheaf you're looking for. If $R$ is not zero dimensional as a ring, there might be games to play.
For your first question in the body, you can beat the first condition by taking an infinite direct sum of sheaves which have nonzero cohomology in some degree since cohomology commutes with direct sums on noetherian schemes. This still won't let you find $H^{n+1}\neq 0$, though, by the theorem referenced above. This also provides a reasonable answer to your second question - finiteness can fail, but all cohomology is concentrated in degrees $0$ through $\dim(X)$. For the third question, take $\bigoplus_{n\in \Bbb Z} \mathcal{O}(n)$.
