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 $X$).

[UMich dissertation][1]

**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.

  [1]: http://deepblue.lib.umich.edu/bitstream/2027.42/60794/1/ofelguei_1.pdf