I think the following *interpretation* of your question is false for cardinality reasons. It occurred to me before I saw Donu's answer but seems to have a similar flavour.

I will say that two curves over a field are *twists of each other* if they become isomorphic over an algebraic closure.

**Interpretation.** Fix a curve $C$ over $\mathbf{Q}$. For every place $v$ of $\mathbf{Q}$, let $X_v$ be an arbitrary (random) twist of $C_v$. Does there exist a curve $D$ over $\mathbf{Q}$ such that $D_v$ is isomorphic to $X_v$ for every $v$ ?

In genus $0$, not every such family can come from a conic over $\mathbf{Q}$, as Felipe has remarked.

In genus $g>0$, if we start with a hyperelliptic curve $C$, then we get *uncountably many* families $(X_v)_v$ by taking random quadratic twists at each place $v$. All these families satisfy your hypotheses, and some cannot come from the *countably many* genus-$g$ curves over $\mathbf{Q}$.

**Examples.** Suppose we have an elliptic curve $E$ over $\mathbf{Q}$ which is the only curve in its isogeny class, and assume moreover that Ш$(E)$ is trivial. It follow that if a genus-$1$ curve $C$ is such that its jacobian $J$ is isomorphic to $E$ almost everywhere locally, then $J$ is $\mathbf{Q}$-isogenous to $E$, and hence $\mathbf{Q}$-isomorphic to $E$, and hence $C$ is isomorphic to $E$ (because Ш$(E)$ is trivial).

Construct the family $(X_v)_v$ by taking $X_v=E_v$ for every place $v\neq2$ of $\mathbf{Q}$, and perversely take $X_2$ to be a quadratic twist of $E_2$, so that $X_2$ and $E_2$ are not $\mathbf{Q}_2$-isomorphic.

If there were a genus-$1$ curve $C$ such that $C_v$ is $\mathbf{Q}_v$-isomorphic to $X_v$ at every $v$, then $C$ would have to be $\mathbf{Q}$-isomorphic to $E$ (by the choice of $E$, as explained above), which is impossible because they are not $\mathbf{Q}_2$-isomorphic.

It remains to find such an $E$. I'm sure this can be done by looking at the tables made by Cremona and Stein. Could someone please confirm this hunch ?