I think this should be called the relatively ample cone and its closure the relatively nef cone. Yes, this already exists and is sensible.
This dissertation might be somewhat useful (although things are done for smooth $X$ instead of singular $X$).
EDIT: With regards to your two conditions (and the comments above), I think this condition is basically factoriality for the singularity. Suppose $x \in X$ is an isolated singularity as you describe and suppose that we are working with only one singular point.
Claim: Suppose that $X$ is factorial, this means that every divisor through $x$ on $X$ is going to be linearly equivalent to a divisor not passing through $x$.
Proof of claim: Suppose that $D > 0$ is an effective divisor passing through $x \in X$ (the effectivity assumption is harmless since we can write $D = D'-D''$ for effective $D'$ and $D''$). Choose a very ample and sufficiently ample Cartier divisor $A$ such that $\Gamma(X, O_X(A - D))$ is globally generated. Choose a global section $s$ generating the stalk $O_{X,x}(A-D)$ (this is possible since $D$ is Cartier at $x$, since $O_{X,x}$ is a UFD). Consider the inclusion $$\Gamma(X, O_X(A-D)) \subseteq \Gamma(X, O_X(A))$$ and so view $s$ as a global section of $O_X(A)$. It follows that $s$ determines a divisor $S \sim A$. Furthermore, $S$ and $D$ coincide in a neighborhood of $x \in X$ by construction. Further choose $T \sim A$ to be any divisor not passing through $x$ (this is possible since $A$ is very ample). Now then $D - S + T \sim D$, but $D - S + T$ is zero near $x \in X$. This proves the claim.
It follows from the claim that there exists a set of generating divisors of the type you want.
On the other hand, if you have divisors on $\hat{X}$ that satisfies those two conditions, then the images (pushforwards) of those divisors also generate the divisor class group of $X$. In particular, the divisor class group of $X$ is generated by divisors not passing through $x$. In particular, it is generated by Cartier divisors (since $x \in X$ was the only singular point). This implies that $X$ is factorial.