For the first question, note that if we assume Dependent Choice then all $\sigma$-additivity arguments can be carried out perfectly. In such setting using the Caratheodory theorem makes perfect sense, and indeed the Lebesgue measure is the completion of the Borel measure.
It was proved by Solovay that $\mathsf{ZF+DC}$ is consistent with "Every set is Lebesgue measurable", relative to an inaccessible cardinal. This shows that assuming large cardinals are not inconsistent, we cannot prove the existence of a non-measurable set (Vitali sets, Bernstein sets, ultrafilters on $\omega$, etc.) without appealing to more than $\mathsf{ZF+DC}$.
Under the assumption of $\mathsf{DC}$ it is almost immediate that there are only continuum Borel sets, but if we agree to remove this assumption then it is consistent that every set is Borel, where the Borel sets are the sets in the $\sigma$-algebra generated by open intervals with rational endpoints. For example in models where the real numbers are a countable union of countable sets; but not only in such models. Do note that in such bizarre models the Borel measure is no longer $\sigma$-additive.
For sets with Borel codes the proof follows as in the usual proof in $\mathsf{ZFC}$. There are only $2^{\aleph_0}$ possible codes, but there are $2^{2^{\aleph_0}}$ subsets of the Cantor set, all of which are codible-Lebesgue measurable.
Lastly, we have a very good definition for the Lebesgue measure, it is simply the completion of the Borel measure. The Borel measure itself is unique, it is the Haar measure of the additive group of the real numbers. The Lebesgue measure is simply the completion of the Borel measure which is also unique by definition. This is similar to the case where the rational numbers could have two non-isomorphic algebraic closures, but there is always a canonical closure. Even if you can find two ways to complete a measure, "the completion" would usually refer to the definable one (adding all subsets of null sets).
Hint: Think about Bass's Proposition 4.14 (1): given any $\delta > 0$, there is an open set $G$ with $A \subset G$ and $m(G - A) < \delta$. Then recall (or prove) that every open set is a countable union of disjoint open intervals.
Here are some more details. An argument like yours has difficulties with the possibility $m(A) = \infty$, but we can reduce to the case $m(A) < \infty$ by intersecting $A$ with large bounded sets, as below.
Suppose $\epsilon \in (0,1)$ is such that $m(A \cap I) \le (1-\epsilon)m(I)$ for each interval $I$. Let $k$ be a positive integer and let $A_k = A \cap [-k,k]$. Note that $m(A_k) \le 2k < \infty$, and that for each interval $I$, we have $$m(A_k \cap I) \le m(A \cap I) \le (1-\epsilon) m(I). \tag{1}$$
Let $\alpha > 0$ be arbitrary, and using Proposition 4.14, choose an open set $G$ with $A_k \subset G$ and
$$m(G - A_k) < \alpha \epsilon. \tag{2}$$
In particular, $m(G) = m(A_k) + m(G-A_k) < 2k + \alpha \epsilon < \infty$.
Now $G$ can be written as a disjoint union of open intervals: $G = \bigcup_{n=1}^\infty I_n$. Note that $m(I_n) \le m(G) < \infty$ for each $n$. Also, $G - A_k = \bigcup_{n=1}^\infty (I_n - A_k)$ which is also a disjoint union.
Now since $I_n - A_k = I_n - (I_n \cap A_k)$, we have
$$m(I_n - A_k) = m(I_n) - m(I_n \cap A_k) \ge m(I_n) - (1-\epsilon)m(I_n) = \epsilon m(I_n) \tag{3}$$ using (1). So by countable additivity,
$$m(G - A_k) = \sum_{n=1}^\infty m(I_n - A_k) \ge \sum_{n=1}^\infty \epsilon m(I_n) = \epsilon m(G). \tag{4}$$
Combining (1) and (3), we get $\epsilon m(G) < \alpha \epsilon$, so $m(G) < \alpha$. In particular, since $A_k \subset G$, we have $m(A_k) < \alpha$. But $\alpha > 0$ was arbitrary, so we must have $m(A_k) = 0$.
Moreover, $k$ was arbitrary, so $m(A \cap [-k,k]) = 0$ for every $k$. Since $A = \bigcup_{k = 1}^\infty (A \cap [-k,k])$, by countable additivity we conclude $m(A) = 0$.
Now let's drop the assumption $m(A) < \infty$. For any $n$ and any interval $I$, we have $$m(A \cap [-n,n] \cap I) \le m(A \cap I) \le (1-\epsilon)m(I).$$
Hence $A \cap [-n,n]$ satisfies the same condition and moreover $m(A \cap [-n,n]) \le 2n < \infty$. So by the previous case, $m(A \cap [-n,n]) = 0$. Now since $A = \bigcup_{n=1}^\infty (A \cap [-n,n])$, by countable additivity $m(A) = 0$.
For an explicit example with a fat Cantor set, let's consider the example given on Wikipedia, where at stage $n \ge 1$ we remove $2^{n-1}$ intervals, each of length $2^{-2n}$. The final set $C$ has measure $1 - \sum_{n=1}^\infty 2^{n-1} \cdot 2^{-2n} = 1 - \sum_{n=1}^\infty 2^{-n-1} = \frac{1}{2}$.
Let $I_k$ be the leftmost interval that remains after stage $k$. At stage $k$ the leftmost interval that was removed was centered at $2^{-k}$ and had length $2^{-2k}$, so the leftmost interval that remains is
$$I_k = \left[0, 2^{-k} - \frac{1}{2} 2^{-2k}\right] = [0, 2^{-k}(1-2^{-k-1})].$$
At the next stage, we will remove from $I_k$ one interval of length $2^{-2(k+1)}$ from $I$, then two intervals of length $2^{-2(k+2)}$ and so on. So the total length of the intervals removed from $I_k$ is
$$\sum_{n=1}^\infty 2^{n-1} 2^{-2(k+n)} = 2^{-2k} \sum_{n=1}^\infty 2^{-n-1} = 2^{-2k} \frac{1}{2} = 2^{-2k-1}.$$
Therefore, we have
$$\frac{m(C \cap I_k)}{m(I_k)} = \frac{m(I_k) - 2^{-2k-1}}{m(I_k)} = \frac{2^{-k}(1-2^{-k-1}) - 2^{-2k-1}}{2^{-k}(1-2^{-k-1})} = \frac{(1-2^{-k-1}) - 2^{-k-1}}{(1-2^{-k-1})}$$
upon cancelling a factor of $2^{-k}$. It is clear by inspection that $$\lim_{k \to \infty} \frac{m(C \cap I_k)}{m(I_k)} = 1,$$ so given $\epsilon > 0$ you can choose $k$ so large that $m(C \cap I_k) > (1-\epsilon) m(I_k)$.
Best Answer
We claim that any measure $\mu$ of your desired form must be the unique measure that maps every countable set to $0$ and every uncountable Borel set to $\infty$.
Consider the sets \begin{align*}C_1&=\{0\le x<1:\textrm{the decimal expansion of $x$ has only digits 0 or 1}\}\\C_2&=\{0\le x<1:\textrm{the decimal expansion of $x$ has only digits 0 or 2}\}\end{align*} Standard "Cantor-set" arguments show that
Now let $B$ be any uncountable Borel set. We'll show that $\mu(B)=\infty$.
We use the so-called perfect set theorem for Borel sets which says that every uncountable Borel set $B$ contains a subset $B'$ homeomorphic to the standard Cantor set $C$. The final item above shows that $B'$ can then be partitioned into uncountably many homeomorphic copies of $C_1$. We write $$B\supseteq B'=\bigsqcup_{i\in I}C^i$$ where each $C^i\cong C_1$ is homeomorphic. By this homeomorphism, each $C^i$ is also uncountable compact, and hence $\mu(C^i)>0$. It follows that $$\mu(B)\ge\sum_{i\in I}\mu(C^i)=\infty$$ since the last sum is an uncountable sum of positive summands.
Addendum, per our previous Discord conversation:
From now on a Borel measure on $\mathbb{R}$ is, as you defined, any measure space $(\mathbb{R},\mathcal{F},\mu)$ where the measurable sets $\mathcal{F}\supseteq\mathbf{B}(\mathbb{R})$ contain all Borel sets and possibly more.
A weird measure is such a measure where the $\mu$-null sets $\textsf{null}_\mu\subseteq\mathcal{F}$ coincide with $[\mathbb{R}]^{\le\aleph_0}\subseteq\mathbf{B}(\mathbb{R})$ the set of all countable subsets of $\mathbb{R}$, and moreover there exists $X\in\mathcal{F}$ measurable with $0<\mu(X)<\infty$.
Sketches.
(1) We prove a stronger statement where $\textsf{CH}$ is replaced with $\mathop{\textsf{add}}(\textsf{null})=\mathfrak{c}$ and in the definition of $\mu$ we replace with $\textsf{null}_\mu=[\mathbb{R}]^{<\mathfrak{c}}$, rewording everything else accordingly (eg. for every instance of "countable" we say "less than continuum" instead). Clearly this implies part (1) of the claim.
Using a usual $\mathop{\textsf{add}}(\textsf{null})=\mathfrak{c}$ construction, take an increasing cofinal sequence $\langle A_i:i<\mathfrak{c}\rangle$ in the poset $(\textsf{null},{\subseteq})$, and let $\langle K_i:i<\mathfrak{c}\rangle$ enumerate all the Lebesgue non-null compact sets. Set $X=\{x_i:i<\mathfrak{c}\}$ with $x_i\in K_i\smallsetminus A_i$ so that $X\cap A\ne\varnothing$ for all Lebesgue non-null $A$ and $\left|X\cap N\right|<\mathfrak{c}$ for all Lebesgue null $N$.
Let $\nu$ be the Gaussian probability measure on $\mathbb{R}$, and restrict $\nu$ to $X$ by setting $\nu^*(X\cap B)=\nu(B)$ for every Borel $B$. Note $\nu^*$ is well-defined with $\nu^*(X)=1$ and $\textsf{null}_{\nu^*}=[X]^{<\mathfrak{c}}$. We let $\mathcal{S}=\mathbf{B}(\mathbb{R})\oplus X$ be the Borel $\sigma$-algebra enriched with $X$, and set $$\mu(A)=\nu^*(A\cap X)+\begin{cases}0,&\left|A\smallsetminus X\right|<\mathfrak{c}\\\infty,&\left|A\smallsetminus X\right|=\mathfrak{c}\end{cases}$$ Then $\mu$ satisfies $\textsf{null}_\mu=[\mathbb{R}]^{<\mathfrak{c}}$, with $\mu(X)=1$ witnessing weirdness. $\square$
(2) It suffices to show every uncountable $X\in\mathcal{F}$ contains a subpartition $X\supseteq\bigsqcup_{i<\omega_1}X_i$ with each $X_i\in\mathcal{F}$ uncountable. Then $\mu(X)\ge\sum_{i<\omega_1}\mu(X_i)=\infty$ works, since $\mu(X_i)>0$ for all $i$.
It's known that $\textsf{MA}_{\aleph_1}$ implies for every $A\sqcup B\subseteq\mathbb{R}$ with $\left|A\right|=\left|B\right|=\aleph_1$, there exists a $G_\delta$ set $A\subseteq C\subseteq\mathbb{R}\smallsetminus B$. My earlier MSE post uses a variation of a proof of this fact.
Take any subpartition $X\supseteq \bigsqcup_{i<\omega_1} A_i$ where each $\left|A_i\right|=\aleph_1$. Take $G_\delta$ sets $\langle C_i:i<\omega_1\rangle$ with $A_i\subseteq C_i\subseteq\mathbb{R}\smallsetminus\bigsqcup_{i<j<\omega_1}A_j$. Then $X_i=X\cap\left(C_i\smallsetminus\bigcup_{j<i}C_j\right)$ works, as $A_i\subseteq X_i$ and $\bigsqcup_{i<\omega_1}X_i\subseteq X$. $\square$
Remark.
Of course you might notice now that asking the same question but with every instance of "countable" replaced with "less than $\mathfrak{c}$" compels one to work in $\textsf{ZFC}+\lnot(\textsf{MA})$ in case (2) (in fact $\mathop{\textsf{add}}(\textsf{null})<\mathfrak{c}$). I have not given thought to what happens then, but a quick gut feeling says large cardinals are possibly relevant.