Your intuition is correct: the map p : L → S¹ is not nullhomotopic.

If p : L → S¹ was nullhomotopic, then it would factor through the universal cover ℝ of S¹.
The inclusion ι : L → S¹ × S¹ would therefore factor through ℝ × S¹.
Let ι' : L → ℝ × S¹ denote a lift of ι, and let L₀ := ι'(L).
The subset L₀ ⊂ ℝ × S¹ is then one of the connected components of the preimage of L in ℝ × S¹.
Let us call the other components L_n for n ∈ ℤ.

Let C₀ ⊂ ℝ × S¹ be a loop in that separates L₀ from L₁.
We may also assume that C₀ is not nullhomotopic in ℝ × S¹ [this still needs a small argument...].

The projection C ⊂ S¹ × S¹ of C₀ is then a loop in the complement of L,
and represents the element (0,1) of ℤ×ℤ = π₁(S¹ × S¹).

Since L lies in the complement of C and L_n → L,
there exists an n such that L_n lies in the complement of C.
This contradicts your assumtion that p : L_n → S¹ is nullhomotopic.

Your intuition is correct: the map p : L → S¹ is not nullhomotopic.

If p : L → S¹ was nullhomotopic, then it would factor through the universal cover ℝ of S¹.
The inclusion ι : L → S¹ × S¹ would therefore factor through ℝ × S¹.
Let ι' : L → ℝ × S¹ denote a lift of ι, and let L₍₀₎ := ι'(L).
The subset L₍₀₎ ⊂ ℝ × S¹ is then one of the connected components of the preimage of L in ℝ × S¹.
Let us call the other components L_(n) for n ∈ ℤ.

Let C₀ ⊂ ℝ × S¹ be a loop in that separates L₍₀₎ from L₍₁₎.
We may also assume that C₀ is not nullhomotopic in ℝ × S¹ [this still needs a small argument...].

The projection C ⊂ S¹ × S¹ of C₀ is then a loop in the complement of L,
and represents the element (0,1) of ℤ×ℤ = π₁(S¹ × S¹).

Since L lies in the complement of C and L_n → L in the Hausdorff metric,
there exists an n such that L_n lies in the complement of C.
This contradicts your assumtion that p : L_n → S¹ is nullhomotopic.