The conjectures could not be true as stated,
due to simple counterexamples such as $3^8+3^8+3^8+2^9=2^8+2^8+3^9$.

One could exclude such constructions by conjecturing,
in the spirit of Schmidt's Subspace Theorem, that:

if $n<d$, and $A_i$ ($1 \leq i \leq n$) are nonzero integers
with $\gcd(A_1,\ldots,A_n)$ such that
$h(|A_i|) \geq d$ for each $i$ and $\sum_{i=1}^n A_i = 0$,
then some proper subsum of the $A_i$ vanishes.

(This accounts for the above "simple counterexample":
$A_i = 3^8, 3^8, 3^8, 2^9, -2^8, -2^8, -3^9$ has
$(n,d)=(7,8)$ but $3^8+3^8+3^8+(-3^9)=0$.)

However, even this refined conjecture is false:
there are has counterexamples with $(n,d) = (5,6)$.
One is $p^6 + q^6 + q^6 + 61^9 r^6 = 2 s^6$ where
$$
\begin{gather}
p \; = \!\! & 37471640786194861459344702995419531,\cr
q \; = \!\! & 20793522547111333210520476761092295,\cr
r \; = \!\! & 3391542261700904858222899444621,\phantom{0000}\cr
s \; = \!\! & 33700711308284627431803214879783946,
\end{gather}
$$
and each of $p^6, q^6, 61^9 r^6, 2 s^6$ has $h=6$
(the last because $s$ is even --
were it not for the single factor of $2$ in $2q^6$,
this identity would have given a counterexample with $(n,d)=(4,6)$.
A similar counterexample, this one with three positive and
two negative $A_i$, is $p^6 + q^6 + q^6 = 11^9 r^6 + 2 s^6$ where
$$
{\small
\begin{gather}
p \; = \!\! & 122143812902307972831486996789219854509652892482229598069
\phantom{0}
\cr
q \; = \!\! & 1754343120851725061884697722096469904639987931170348892227
\cr
r \; = \!\! & 53451023851036429085688858950495539530964060758748930439
\phantom{00}
\cr
s \; = \!\! & 1088043146197825196095684124547610617079707688400198829578.
\end{gather}
}
$$

Both of these solutions were obtained using the identity
$$
(q^2+qs-s^2)^3 + (q^2-qs-s^2)^3 = 2(q^6-s^6).
$$
(This identity is not new; Dickson's *History of the Theory of Numbers, Vol. II*
attributes an equivalent identity to Gérardin in 1910, see page 562 note 107.)
We cannot nontrivially make both of $|q^2 \pm qs - s^2|$ squares, because
that yields elliptic curves of rank zero. But we can make one of them $p^2$
and the other $\delta r_1^2$ for some choices of $\delta$ that yield
elliptic curves $E$ of positive rank, and then search the group of rational
points for examples with $\delta | r_1$ (so we can use $r = r_1 / \delta$
and obtain solutions of $p^6 \pm \delta^9 r^6 = 2(q^6-s^6)$).
The first such $\delta$ is $11$, with $(q,s) = (3,-2)$ making
$q^2+qs-s^2 = -1$ and $q^2-qs-s^2 = 11$.
One must multiply the generator by $11$ to get $11|r_1$;
that's how I found the second example. The first has $\delta = 61$,
using an elliptic curve of rank $2$ with independent solutions
$(q,s) = (10,3)$ and $(26,15)$; while these are more complicated than
the $\delta = 11$ generator, and $61 | r_1$ is harder to get than $11 | r_1$,
we still end up with a smaller example thanks to the freedom to choose
two multipliers $-$ the one above uses multipliers $4$ and $5$ respectively.