I'll prove that the modified question 1 has an affirmative answer by showing conversely that an infinite word $u\in\{0,1\}^{\mathbb Z}$ which avoids all $0w01\widetilde w1 $ and $1w10\widetilde w0 $ (call these factors *forbidden*) is necessarily balanced.

If $u\ne...0101...$, there is wlog a factor 00. (Otherwise, swap all 0's and 1's). For $k\ge1$, define a *$k$-patch* (or $p_k$ for short) as a 'maximal' run of $k$ consecutive zeros in $u$, i.e. it is preceded and followed by a 1. As 0011 and 1100 are forbidden, these 1's are isolated if $k\ge 2$, so each $10_k1$ (where indices mean repetition) is preceded and followed by patches $p_a$ and $p_b$, and the forbidden factors imply $k-1\le a,b\le k+1$. The same holds trivially for $k=1$. So **two 'neighboring' patches have a difference of length of at most 1.**

Now suppose that there is a $k\ge 1$ such that $u$ contains a $p_{k-1}$ and a $p_{k+1}$. (It makes sense to define a $0$-patch $p_0$ as the empty set in the middle of a factor 11 forcing those two 1's.)
Suppose they are closest possible to each other, i.e. between them there is only a number, say $n$, of $k$-patches (each separated by an isolated 1).

So $u$ has a factor $$ 10_{k-1}\overbrace{10_k\cdots10_k}^n10_{k+1}1$$ (or the other way round, in which case inverse the whole sequence wlog). Which entries must follow to the right? In fact we must have

$$ 10_{k-1}\overbrace{10_k\cdots10_k}^n\underbrace{10}_{(x)}0_{k-1} \underbrace{01}_{(y)}\overbrace{\color{red}{0_k1\cdots0_k1}}^{n-1}\color{red}{0_k}$$ because each red 0 is forced by the fact that the string 01 marked by $(y)$ must not be the center of a forbidden factor, and each red 1 is forced by the fact that the string 10 marked by $(x)$ must not be the center of a forbidden factor. But due to the initial 1, the whole string above is a forbidden one with center $(x)$. Contradiction.

So there is a $k$ such that $u$ consists only of patches of lengths $k$ and $k+1$, separated by isolated 1's. ~~It is easy to see that such a $u$ is balanced. qed.~~

**EDIT**: this was wrong, but the proof can be rescued by the following completion.

Suppose $u$ is not balanced. Then take the smallest $r$ such that there are two factors $s$ and $t$ of length $r$ where $t$ has 2 more 1's than $s$. So $s$ starts and ends with $0_{k+1}$ and $t$ starts and ends with $10_k1$.

Let $n$ be the number of interior 1's of each $s$ and $t$. Suppose there are $i$ instances of $p_k$ and $j$ instances of $p_{k+1}$ with both $i,j>0$, then the interior of $t$ must have the same numbers of each (note that $s$ and $t$ have the same number of 1's in their interior).

**EDIT after the last remark of Harry Altman** (the one starting with "Yay") : I have found a way to repare the proof again, making it at the same time more elegant.

We'll now introduce the process of *reduction*: put $a^0=1,b^0=0$ and $u^0=u$. For $\nu=0,1,...$ proceed as follows: As the $a^\nu$-$b^\nu$-word $u^\nu$ contains no forbidden factors (in terms of $a^\nu$ and $b^\nu$), the boldface statement above about 'neighboring' patches applies also to $u^\nu$, so there must be a $k^\nu$ such that $u^\nu$ is composed either

(1) of runs $a^\nu_{k^\nu}$ and $a^\nu_{k^\nu+1}$, interspearsed by isolated $b^\nu$'s, or

(2) of runs $b^\nu_{k^\nu}$ and $b^\nu_{k^\nu+1}$, interspearsed by isolated $a^\nu$'s.

Then denote in case (1) $a^{\nu+1}:=b^\nu a^\nu_{k^\nu}, b^{\nu+1}:=b^\nu a^\nu_{k^\nu+1}$, and

in case (2) $a^{\nu+1}:=a^\nu b^\nu_{k^\nu}, b^{\nu+1}:=a^\nu b^\nu_{k^\nu+1}$,

further define $u^{\nu+1}$ as the sequence $u$ written as a $a^{\nu+1}$-$b^{\nu+1}$-word. This reduction can be traced back, and it is easy to see that that $u^{\nu+1}$ cannot contain $a^{\nu+1}$-$b^{\nu+1}$-words as forbidden factors. (Looking at the $a^{\nu+1}b^{\nu+1}$ or $b^{\nu+1}a^{\nu+1}$ in the center of a forbidden factor, it is clear that this gives rise to a forbidden factor in $u^\nu$, which is excluded.)

$s^\nu$ and $t^\nu$ can be written as factors of $u^\nu$, one of them starts and ends with $a^\nu$’s, the other one of them starts and ends with $b^\nu$’s. By the minimality of $s$ and $t$ and the fact that they have the same number of letters, it is easy to see that, up to an initial or final isolated letter (like in the $\nu=0$ case : a missing 1 before $s$ and the final 1of $t$), they must factor also in terms of the ‘new letters’ $a^{\nu+1}$’s and $b^{\nu+1}$, and their numbers of occurrences in the interior must again coincide. (To illustrate this, think of $s=bbbabbbabbabbabbb$ and $t=abbabbabbabbbabba$ omitting the $\nu$’s. Here $k=2$, as there are runs in $b$ of lengths 2 and 3.. Then the reduction would be $abbb\to b$ and $abb\to a$ with $s$ (or rather $as$) $\to bbaab$ ant $t$ (or rather $t$ with the final $a$ omitted) $\to aaaba$.)

It is clear that $s^{\nu+1}$ has less letters $a^{\nu+1}$ and $b^{\nu+1}$ than $s^\nu$ has letters $a^\nu$ and $b^\nu$, and that by minimality, one of them starts and ends with $a^{\nu+1}$’s, the other one of them starts and ends with $b^{\nu+1}$’s. Iterating this reduction, we must thus come to a point where (omitting again the $\nu$’s) either $s$ or $t$, say $s$, consists only of letters of the same kind, say $a$. But at that moment, $t$ must have the form $baa\cdots ab$, thus it has a run of two less $a$’s. Contradiction.

We conclude that the initial $s$ and $t$ cannot exist, thus $u$ is balanced. Qed.