Given $S_n = 3^n + r$, and $S_{n-1} = 3^{n-1} + r$, where $n \geq 2$, $n \in \mathbb{N}^+$,

we can find the $n^{th}$ term of the sequence by subtracting the sum of the first $n-1$ terms from the sum of the first $n$ terms:

$$a_n = S_n - S_{n-1} = 3^n + r - (3^{n-1} + r) = 3^n - 3^{n-1} = 2 \cdot 3^{n-1}.$$

Next, we consider the first term of the sequence:

$$a_1 = S_1 = 3 + r.$$

Also from the general term found earlier, $a_2 = 2 \cdot 3^1 = 6$, and since $a_2 = 3 \times a_1$, the common ratio of the sequence is $3$. Therefore:

$$a_1 = \frac{a_2}{3} = \frac{6}{3} = 2.$$

Hence, by comparing it with $a_1 = S_1$, we get:

$$2 = 3 + r,$$

and solving for $r$ gives:

$$r = 2 - 3 = -1.$$

Therefore, the correct answer is $\boxed{r = -1}$, option B.