Unfortunately almost never (although it obviously holds if $j^* \mathcal{F} = 0$). In particular, it virtually never holds for $\mathcal{F} = \mathcal{O}_X$. I'm going to assume that $\mathcal{F}$ is coherent.

Let me give the quick answer first, then I'll explain it in more detail.

**Quick answer:** $R^i j_* j^* \mathcal{F}$ coincides with $H^{i+1}_{X \setminus U}(\mathcal{F})$ the $i+1$th cohomology with support at $Z = X \setminus U$ of $\mathcal{F}$. By basic facts about local cohomology, you can basically never expect this to vanish for all $i > 0$.

**Detailed answer:** Let's suppose that $X = \text{Spec} R$ for simplicity and that $U = X \setminus V(I)$ for some ideal $I$. Further assume that $j^* M = M|_U \neq 0$. Let $m$ be a minimal prime of $I$ such that $M_m \neq 0$. We now can replace $R$ by $R_m$ and $M$ by $M_m$.

Working locally, for any $R$-module $M$ we can identify
$$
R^i j_* j^* M = R^i j_* (M|_U) = H^i(U, M|_U) = H^{i+1}_m(M)
$$
for $i > 0$ where $H^*_m$ is local cohomology. This comes from the long exact sequence
$$
H^i_m(M) \to H^i(X, M) = 0 \to H^i(U, M|_U) \to H^{i+1}_m(M) \to H^{i+1}(X, M) = 0.
$$

Now, $H^{i+1}_m(M) \neq 0$ for any $i$ such that $\dim \text{Supp} M = i+1$ by Theorem 3.5.7 in *Cohen-Macaulay Rings* by Bruns and Herzog.

In particular, if $\dim \text{Supp} M \geq 2$, then we have non-vanishing $R^i j_* j^* M$. Now, if $\dim \text{Supp} M = 1$ you can still do something similar (the dimension can't be zero by our previous assumptions).

In this case, it follows that $H^1_m(M) \neq 0$ and in fact is infinitely generated. Thus $H^0(U, M|_U)$ is also infinitely generated since we have the exact sequence
$$
M \to H^0(U, M|_U) \to H^1_m(M).
$$