I'll write $\longrightarrow$ for some standard iterations and $\implies$ for some possibly non-standard iterations.  The relaxed Collatz conjecture is that for all $x$, $x \implies 1$.  I'll call counterexamples to the conjecture <i>escapees</i>.

First of all, since it's a good example of the notation, a lemma: for all $a$, $4 a \implies 9 a + 1$.

Proof: $4a \implies 12 a + 1 \longrightarrow 36a + 4 \longrightarrow 18a + 2 \longrightarrow 9 a + 1$.

$\square$

Now here I what I will try to show:

Theorem: If $x$ is the smallest escapee, then $4x \implies 1$.

Proof: By elementary considerations, $x \equiv 3 \pmod{4}$.  If $x \equiv 2 \pmod{3}$, then $\frac{2 x - 1}{3} \longrightarrow 2 x \longrightarrow x$.  So in this case, $2 x$ can't be an escapee, and neither can $4 x$.

The next case is $x \equiv 0 \pmod{3}$.  Write $x = 12k+3$.  Using the lemma, $4x \implies 27k+7$.  Observe that $x > 9k+2 \rightarrow 27k+7$ if $k$ is odd.  Now if $x \equiv 15 \pmod{24}$, we also have that $4x \implies 1$.

It's also possible that $x \equiv 3 \pmod{24}$.  In that case, we have $x = 24 j + 3$ and $x \longrightarrow 27j+4$.  This means $j$ cannot be even.  So we drill down again, with $x = 48i + 27 \longrightarrow 81i+49$.  This means $i$ cannot be odd.  But also, $4x = 4(48i+27) \implies 81i + 92$, so if $i$ is even, the next step brings us under $x$.  This covers all the $x \equiv 0 \pmod{3}$ cases.

So far, our result is that if $x \implies 4x$, then $x \equiv 7 \pmod{12}$.  Now I'll handle that case.  Suppose $x \implies 4x$ and $x = 24k + 7$.  Then we have:

$4x = 4(24k+7) \implies 9(24k+7)+1 = 216k+64 \longrightarrow 108k+32 \longrightarrow 54k+16$.

But also, $18k+5 \longrightarrow 54k+16$.  This means $18k+5$ is an escapee too, but it can't be, since it's less than $x$.  So the hypothesis is false and we have $x = 24k + 19$ instead.

Now, consider:

$4x = 4(24k+19) \implies 216k + 172 \longrightarrow 108k + 86 \longrightarrow 54k + 43 \longrightarrow 162k + 130 \longrightarrow 81k+65$.

By similar reasoning, $k \neq 3 \pmod{4}$, or else we can continue the chain two more steps to get an escapee less than $x$.  Finally,

$x = 24k+19 \longrightarrow 72k+58 \longrightarrow  36k + 29 \longrightarrow 108k + 88 \longrightarrow 54k + 44 \longrightarrow 27k + 22$.

shows that $k$ can't be even, and substituting $k = 4 j + 1$:

$27k + 22 = 108j + 49 \longrightarrow 324j + 148 \longrightarrow 162j + 74 \longrightarrow 81j + 37 < x$

completes the proof, since that's all the cases. 

$\square$

Corollary: if the Collatz conjecture is false and the smallest counterexample leads back to itself then that number is not a counterexample to the relaxed Collatz conjecture.  In other words, if $y$ is the smallest number satisfying $\neg(y \longrightarrow 1)$, then $y \longrightarrow y$ implies $y \implies 1$.  This is a weaker version of what I originally thought I had proved.

Corollary: If $4x$ is an escapee, then there is an escapee smaller than $x$.  This suggests to me a recursive descent strategy where we attempt to show $z \implies 4k < 4x$.  But it's not clear if number-crunching will get us anywhere further.