Yes, this is true, as a consequence of an identity in a space of
modular forms of weight $6$.

The form $\eta(2z)^{12}$ is in this space; and $\sigma_5(n)$ for $n$ odd
are the coefficients of the weight-$6$ form
$$
\frac1{1008} \Bigl(E_6(z+\frac12) - E_6(z)\Bigr) =
q + 244 q^3 + 3126 q^5 + 16808 q^7 + 59293 q^9 + \cdots.
$$
The difference is
$$
256(q^3 + 12 q^5 + 66 q^7 + 232 q^9 + 627 q^{11} + 1452 q^{13} + \cdots);
$$
after a bit of experimentation we recognize this as $256q^3$ times the
$12$-th power of $1+q^2+q^6+q^{12}+q^{20}+\cdots$, which is to say
$256$ times the $12$-th power of the weight-$1/2$ modular form
$\sum_{k=0}^\infty q^{(2k+1)^2/4}$. Such a formula, once surmised,
can be proved by comparing initial segments of the $q$-expansions;
I did this to $O(q^{61})$, which is way more than enough.
The congruence mod $256$ follows because the $12$-th power of
$1+q^2+q^6+q^{12}+q^{20}+\cdots$ clearly has integral coefficients.

*Added later*:
This identity (and thus the congruence that **Tito Piezas III** asked for)
gives a formula $(\sigma_5(n) - \rho(n)) / 256$ for the number of
representations of $4n$ as the sum of $12$ odd squares, or equivalently of
$(n-3)/2$ as the sum of $12$ triangular numbers. Following this lead,
I soon found both the formula and the congruence in the paper

Ken Ono, Sinai Robins, and Patrick T. Wahl: On the representation of integers as sums of triangular numbers, *Aequationes Math.* **50** (1995) #1, 73-94

available
on
Ken Ono's page,
where the enumeration is given (in the triangular-number form) as Theorem 7,
and the congruence as a "simple consequence" of that theorem.