Let $X$ be a scheme, and $F$ a quasi-coherent sheaf of $\mathcal{O}_X$-modules on $X$, such that for some dense open $U\subset X$, we have $F|_U = 0$.
Under these assumptions, must $F$ be a torsion module?
I'm happy to assume $X$ is noetherian and reduced if necessary.
Apologies in advance if this post is useless and/or incorrect. Here are some thoughts on what happens if you either assume that $X$ is integral, or assume that $X$ is reduced and noetherian.
(i) ($X$ integral.) Pick an arbitrary open affine subset ${\rm Spec}A \subset X$, where the restriction of $\mathcal F$ is of the form $\widetilde M$ for some $A$-module $M$. Since $U$ is dense, $U$ contains the generic point of $X$. The stalk of $\mathcal F$ at the generic point is the localisation $M_{(0)}$. Since $\mathcal F|_U = 0$, we deduce that $M_{(0)}$ is the zero module. Therefore, for every $m \in M$, there exists a non-zero $r \in A$ such that $rm = 0$. This proves that $M$ is a torsion module. The stalks of $\mathcal F$ inside ${\rm Spec}A$ are localisations of $M$, so these are also torsion modules. Since ${\rm Spec} A$ was chosen arbitrarily, this shows that $\mathcal F$ is a torsion sheaf.
(ii) ($X$ reduced and noetherian.) Again, pick an arbitrary open affine subset ${\rm Spec}A \subset X$, where $\mathcal F$ restricts to $\widetilde M$. Since $A$ is a noetherian ring, it has finitely many minimal prime ideals, $\mathfrak p_1, \dots,\mathfrak p_n$. The open set $U \cap {\rm Spec} A$ is equal to ${\rm Spec}A \ \backslash \ V(I)$ for some ideal $I \subset A$. Clearly $I$ is not a subset of any $\mathfrak p_i$, for if $I \subset \mathfrak p_i$, then $U \cap {\rm Spec} A$ would fail to intersect the non-empty open set ${\rm Spec} A \ \backslash \left( \cup_{j \neq i} V(\mathfrak p_j)\right)$ (because every prime ideal in $A$ contains at least one minimal prime ideal) and this would contradict the fact that $U$ is dense. By the prime avoidance lemma, it follows that $I$ is not a subset of $\cup_i \mathfrak p_i$. Therefore, there exists at least one element $f \notin \cup_i \mathfrak p_i$ that is contained in $I$, i.e. there exists at least one element $f \notin \cup_i \mathfrak p_i$ such that the basic open affine subset $D(f) \subset {\rm Spec} A$ is contained in $U$.
Since $\mathcal F|_U = 0$, it follows that the localisation $M_f = \mathcal F(D(f))$ is the zero module, which implies that for every $m \in M$, there exists an $e \geq 0$ such that $f^e m = 0$. The element $f^e$ cannot be contained in $\cup_i \mathfrak p_i$, otherwise $f$ itself would be contained in $\cup_i \mathfrak p_i$, contrary to assumption. Since $A$ is reduced, $\cup_i \mathfrak p_i$ is precisely the set of zero divisors in $A$. Thus we have shown that, for every $m \in M$, there exists a non-zero-divisor $r \in A$ such that $rm = 0$, which proves that $\mathcal F({\rm Spec}A) = M$ is a torsion module.
Moreover, if $\mathfrak p$ is any point in ${\rm Spec}A$, then the ideals in $A_{\mathfrak p}$ are in one-to-one correspondence with the ideals in $A$ contained inside $\mathfrak p$, so the minimal ideals in $A_{\mathfrak p}$ are the localisations of those $\mathfrak p_i$'s that are contained inside $\mathfrak p$, and the set of zero divisors in $A_{\mathfrak p}$ is the union of the localisations of these $\mathfrak p_i$'s. Thus any $r \in A$ that is a non-zero-divisor in $A$ is also a non-zero-divisor in $A_{\mathfrak p}$. Since every element of $A_{\mathfrak p}$ is of the form $m /g$ for some $m \in M$ and $g \in A \backslash \mathfrak p$, and since there exists a non-zero-divisor $r \in A$ (which is also a non-zero-divisor in $A_{\mathfrak p}$) that annihilates $m$, it follows that the stalk $\mathcal F_{\mathfrak p} = M_{\mathfrak p}$ is a torsion module, and hence, $\mathcal F$ is a torsion sheaf.
I think you are searching for the following:
If $X$ is a noetherian (whether $X$ is reduced is not relevant) scheme, $F_{|U}=0$ and $I$ the ideal sheaf for $X\setminus U$ (say we choose the reduced scheme structure), then $I^nF=0$ for some $n>0$. The proof of this fact is very easy.
If furthermore $U$ happens to be dense, then one can show that $I^n$ contains a regular element (That is what Kenny Wong did in the second part of his answer), i.e. on all affine pieces any element of $F$ is torsion in the sense of commutative algebra.
