Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false.  Note that $a+b$ must be odd.  Put $\ell = a+b$  (and $\ell$ is odd) and $m=a-b$ which is also odd.  The condition that $b<a<2b$ now translates to $0<m < \ell/3$.  The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.  

So the question really is can $m^3-2$ have a square factor of size exceeding $(3m)^2$?   Here's a heuristic that indicates why this should be possible.  Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$.  Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev).  For $p=3$ one should work a tiny bit harder.  Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$.  We may expect that probability to be about $f(\ell)/(3\ell)$.  Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about 
$$ 
\sum_{\ell < x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x,
$$ 
for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$.   This suggests that there are infinitely many counterexamples.  

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.  

Added:  A relevant paper for such problems is Granville's [paper][1] in IMRN.  
Granville shows that on the abc conjecture the largest square factor of any cubic polynomial (without repeated roots) $f(m)$ is at most $m^{2+o(1)}$.  This generalizes the observation in GH from MO's answer (the general case requires more work than the easy $x^3-2$ example).  Granville also gives reasons to believe that this exponent is best possible, and in particular shows that such large square factors are attained for cubic polynomials having a rational root -- this uses the solutions to Pell's equation.  Of course, Granville's construction doesn't apply for $x^3-2$ which is irreducible, but the results may be suggestive.  Also note that the Pell type constructions produce a logarithmic number of examples with large square factors, which is again suggestive.

[1]:   http://www.dms.umontreal.ca/~andrew/PDF/hyperelliptics.pdf