The answer is "no", two generically equal sections are equal.

First prove that the image of a morphism $f:\mathrm{Spec}\,\mathbb{Z}\rightarrow X$ can not intersect both origins. Since $\mathrm{Spec}\,\mathbb{Z}$ is the terminal object in the category of schemes, any morphism $\mathrm{Spec}\,\mathbb{Z}\rightarrow X$ is a section (in particular, a locally closed immersion).

Note that $\mathrm{Spec}\,\mathbb{Z}$ is irreducible, the image of an irreducible space is irreducible and the closure of an irreducible subset is irreducible–so the closure $Y$ of $f(\mathrm{Spec}\,\mathbb{Z})$ is irreducible. It can not have Krull dimension 0 (because $\mathrm{Spec}\,\mathbb{Z}$ is one-dimensional) and it can not have dimension 2 (a closed irreducible subset of top dimension has to be the whole space; because both $\mathrm{Spec}\,\mathbb{Z}$ and $\mathrm{Spec}\,\mathbb{Z}[x]$ are reduced, this would mean that the former is an open subscheme of the latter–but this is not true because function fields are not isomorphic). Therefore, $Y$ is a closed irreducible subscheme of dimension 1. It also has codimension 1 in $X$ because $X$ is a catenary space.

The intersection of $Y$ with a standard affine open $U_1\subset X$ is a non-empty closed subscheme of $U_1$. It is irreducible beucase it can also be considered as a non-empty open subscheme of $Y$. It has dimension 1 because codimension is preserved under taking intersection with opens and $U_1$ is catenary. This means that it is cut out by a prime ideal of height 1; such a prime ideal is generated either by a prime number $p$ or a non-zero irreducible polynomial $f(x)$. It can not be the first because $f(\mathrm{Spec}\,\mathbb{Z})\approx \mathrm{Spec}\,\mathbb{Z}$ is an open subscheme of $Y$ so its function field has to be $\mathbb{Q}$; it can not be the second because $Y$ then could not intersect both $x=0$ and $x\neq 0$ (and it has to intersect the latter since otherwise $f(\mathrm{Spec}\,\mathbb{Z})$ would be reducible). Contradiction.

So the image of any morphism $\mathrm{Spec}\,\mathbb{Z}\rightarrow X$ has to lie in one of the two standard affine opens. It is clear that generically equal morphisms $\mathrm{Spec}\,\mathbb{Z}\rightarrow \mathrm{Spec}\,\mathbb{Z}[x]$ are equal.

EDIT: since there are many down-votes and no comments, I would guess that the last step seems unjustified (if this edit seems to be useless, feel free to down-vote but please leave a comment).

Recall that a morphism $\mathrm{Spec}\,\mathbb{Z}\rightarrow \mathrm{Spec}\,\mathbb{Z}[x]$ is equivalent to the data of a morphism of commutative unital rings $\mathbb{Z}[x]\rightarrow \mathbb{Z}$. If two morphisms $f_1$, $f_2:\mathbb{Z}[x]\rightarrow \mathbb{Z}$ define generically equal sections, then $f_1(a)\otimes b-f_2(a)\otimes b=0\in \mathbb{Q}$ for all $a\in \mathbb{Z}[x], b \in \mathbb{Q}$. By linearity of tensor product, this implies $(f_1(a)-f_2(a))\otimes b=0$ for all $a\in \mathbb{Z}[x], b \in \mathbb{Q}$. But there is no non-zero $c\in \mathbb{Z}$ such that $c\otimes b=0$ for some non-zero $b \in \mathbb{Q}$. Therefore, $f_1(a)-f_2(a)=0$, i.e. two morphisms are equal.