The Krull dimension, as defined by Gabriel and Rentschler, of not-necessarily commutative rings is an ordinal. See, for example, [John C. McConnell, James Christopher Robson, Lance W. Small, Noncommutative Noetherian rings].
More generally, they define the deviation of a poset $A$ as follows. If $A$ does not have comparable elements, $\mathrm{dev}\;A=-\infty$; if $A$ is has comparable elements but satisfies the d.c.c., then $\mathrm{dev}\;A=0$. In general, if $\alpha$ is an ordinal, we say that $\mathrm{dev}\;A=\alpha$ if (i) the deviation of $A$ is not an ordinal strictly less that $\alpha$, and (ii) in any descending sequence of elements in $A$ all but finitely many factors (ie, the intervals of $A$ determined by the successive elements in the sequence) have deviation less that $\alpha$.
Then the Gabriel-Rentschler left Krull dimension $\mathcal K(R)$ of a ring $R$ is the deviation of the poset of left ideals of $R$. A poset does not necessarily have a deviation, but if $R$ is left nötherian, then $\mathcal K(R)$ is defined.
A few examples: if a ring is nötherian commutative (or more generally satisfies a polynomial identity), then its G-R Krull dimension coincides with the combinatorial dimension of its prime spectrum, so in this definition extends classical one when these dimensions are finite. A non commutative example is the Weyl algebra $A_{n}(k)$: if $k$ has characteristic zero, then $\mathcal K(A_n(k))=n$, and if $k$ has positive characteristic, $\mathcal K(A_n(k))=2n$. The book by McConnel and Robson has lots of information and references.
In the setup in the question, it should really say "we could have invertible meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$", since those are what give rise to "extra principal Cartier divisors". This is what I will prove cannot happen. The argument is a correction on an earlier attempt which had a bone-headed error. [Kleiman's construction from Georges' answer is not invertible, so no inconsistency. Kleiman makes some unfortunate typos -- his $\oplus k(Q)$ should be $\prod k(Q)$, and more seriously the $t$ at the end of his construction should be $\tau$, for example -- but not a big nuisance.]
For anyone curious about general background on meromorphic functions on arbitrary schemes, see EGA IV$_4$, sec. 20, esp. 20.1.3, 20.1.4. (There is a little subtle error: in (20.1.3), $\Gamma(U,\mathcal{S})$ should consist of locally regular sections of $O_X$; this is the issue in the Kleiman reference mentioned by Georges. The content of EGA works just fine upon making that little correction. There are more hilarious errors elsewhere in IV$_4$, all correctable, such as fractions with infinite numerator and denominator, but that's a story for another day.) Also, 20.2.12 there is the result cited from Qing Liu's book in the setup for the question.
The first step in the proof is the observation that for any scheme $X$, the ring $M(X)$ of meromorphic functions is naturally identified with the direct limit of the modules Hom($J, O_X)$ as $J$ varies through quasi-coherent ideals which contain a regular section of $O_X$ Zariski-locally on $X$. Basically, such $J$ are precisely the quasi-coherent "ideals of denominators" of global meromorphic functions. This description of $M(X)$ is left to the reader as an exercise, or see section 2 of the paper "Moishezon spaces in rigid-analytic geometry" on my webpage for the solution, given there in the rigid-analytic case but by methods which are perfectly general.
Now working on Spec($A$), a global meromorphic function "is" an $A$-linear map $f:J \rightarrow A$ for an ideal $J$ that contains a non-zero-divisor Zariski-locally on $A$.
Assume $f$ is an invertible meromorphic function: there are finitely many $s_i \in J$ and a finite open cover {$U_i$} of Spec($A$) (yes, same index set) so that $s_i$ and $f(s_i)$ are non-zero-divisors on $U_i$; we may and do assume each $U_i$ is quasi-compact. Let $S$ be the non-zero-divisors in $A$. Hypotheses are preserved by $S$-localizing, and it suffices to solve after such localization (exercise). So without loss of generality each element of $A$ is either a zero-divisor or a unit. If $J=A$ then $f(x)=ax$ for some $a \in A$, so $a s_i=f(s_i)$ on each $U_i$, so all $a|_{U_i}$ are regular, so $a$ is not a zero-divisor in $A$, so $a$ is a unit in $A$ (due to the special properties we have arranged for $A$). Hence, it suffices to show $J=A$.
Since the zero scheme $V({\rm{Ann}}(s_i))$ is disjoint from $U_i$ (as $s_i|_ {U_i}$ is a regular section), the closed sets $V({\rm{Ann}}(s_i))$ and $V({\rm{Ann}}(s_2))$ have
intersection disjoint from $U_1 \cup U_2$. In other words, the quasi-coherent ideals ${\rm{Ann}}(s_1)$ and ${\rm{Ann}}(s_2)$ generate the unit ideal over $U_1 \cup U_2$.
A quasi-coherent sheaf is generated by global sections over any quasi-affine scheme, such as $U_1 \cup U_2$ (a quasi-compact open in an affine scheme), so we get $a_1 \in {\rm{Ann}}(s_1)$ and $a_2 \in {\rm{Ann}}(s_2)$ such that $a_1 + a_2 = 1$ on $U_1 \cup U_2$. Multiplying both sides by $s_1 s_2$, we get that $s_1 s_2 = 0$ on $U_1 \cup U_2$. But
$s_1$ is a regular section over $U_1$, so $s_2|_ {U_1} = 0$. But $s_2|_ {U_2}$ is a regular section, so we conclude that $U_1$ and $U_2$ are disjoint. This argument shows that the $U_i$ are pairwise disjoint.
Thus, {$U_i$} is a finite disjoint open cover of Spec($A$), so in fact each $U_i = {\rm{Spec}}(A_i)$ with $A = \prod A_i$. But recall that in $A$ every non-unit is a zero-divisor. It follows that the same holds for each $A_i$ (by inserting 1's in the other factor rings), so each regular section $s_i|_ {U_i} \in A_i$ is a unit. But the preceding argument likewise shows that $s_i|_ {U_j} = 0$ in $A_j$ for $j \ne i$, so each $s_i \in A$ has a unit component along the $i$th factor and vanishing component along the other factors. Hence, the $s_i$ generate 1, so $J = A$. QED
Best Answer
The dimension of $R[1/v]$ is the biggest height of some prime ideal $P$ such that $v\notin P$. So, let $I_{d-1}$ be the intersection of all primes of height at least $d-1$ ($d= \dim R$), then
Under a mild condition (all maximal ideals has height at least $d-1$), then $I_{d-1}$ is inside the Jacobson radical of $R$. So in this case $v$ is not inside the Jacobson radical would suffice.
EDIT (in response to the comment by the OP): Now let $I_d$ be the intersection of all primes with height $d$. Then $\dim R[1/v] \leq d-1$ iff $v\in I_d$. So
Here is a short calculation to show that in Fernando's example above, $I_1=(x)$. Any prime that does not contain $x$ would have to contain $(x-1, y, z)$, but this is a maximal ideal of height $0$. So $x \in I_1$. On the other hand $R/(x)$ is $k[y,z]$ so the intersection of all height one there is $0$. Thus $I_1=(x)$. This gives you precisely what elements satisfies your condition.