No – the statement you would like to have is already false in the very smallest non-trivial example.
We exhibit a single cyclic module for which

• the Littlewood–Richardson coefficient
c^{,n\lambda}{,n\mu , n\nu} = c^{(2n)}{(n),(n)} = 1

is non–zero,
• but no submodule (L) of type (\lambda) has the required behaviour with respect to a given submodule (N) of type (n\mu).

(The same construction works for any discrete valuation ring; to avoid notation write (\mathfrak o=\mathbb Z_{p}) and (\pi=p).)

The data

Take
 λ = (2), μ = (1), ν = (1) and n ≥ 2 .

1. The module
   M := (\mathfrak o/\pi^{2n})
   has type (n\lambda = (2n)).

2. The submodule
   N := (\pi^{,n}M) ≅ (\mathfrak o/\pi^{n})
   has type (n\mu = (n)).

3. The quotient
   M∕N ≅ (\mathfrak o/\pi^{n})
   has type (n\nu = (n)).

So the hypotheses you ask for are satisfied.

Why no suitable (L) exists

Because (M) is cyclic, every submodule is again cyclic; more precisely

 for 0 ≤ s ≤ 2n, π^{s}M ≅ (\mathfrak o/\pi^{,2n-s}).

Thus a submodule has type (length) 2 exactly when its length is 2, i.e. when
s = 2n − 2. In other words

 the unique submodule of type λ = (2) is
  L₀ := π^{,2n-2}M.

Consequently every candidate (L) must equal (L_{0}).

Now

 L₀ ∩ N = π^{,\max{2n-2,,n}}M = π^{,2n-2}M = L₀,

because n ≤ 2n − 2 for every n ≥ 2. Therefore

 type(L₀ ∩ N) = (2),

whereas we would need it to be μ = (1). Hence no submodule (L) with the required properties exists.

Conclusion

Even though
c^{,n\lambda}{,n\mu , n\nu} ≠ 0  (implying c^{\lambda}{\mu , \nu} ≠ 0 by Knutson–Tao), one cannot, in general, ``divide by (n)’’ inside a fixed module to obtain a submodule (L) realising the smaller Littlewood–Richardson data. The strengthened version of the Saturation Conjecture therefore fails.