To make the question in the title precise, let me phrase it like this. Consider a complete local ring

$$ A := \mathbb{C}[[x_1, \dotsc, x_n]]/(f_1, \dotsc, f_m) $$

and, for the sake of definiteness, assume that $A$ is (S$_2$) (although I bet that this will be irrelevant). The goal is to deform $A$ to a much nicer ring, whose singularities are significantly better. To have a chance at a positive answer, the deformations we consider are in a very weak, explicit sense: for elements $g_1, \dotsc, g_m \in \mathbb{C}[[x_1, \dotsc, x_n]]$ that will be allowed to be chosen freely, we consider the rings

$$ B_{g_1, \dotsc, g_m} := \mathbb{C}[[x_1, \dotsc, x_n, y_1, \dotsc, y_m]]/(f_1 + y_1g_1, \dotsc, f_m + y_mg_m) $$

and

$$ B'_{g_1, \dotsc, g_m} := \mathbb{C}[[x_1, \dotsc, x_n, y]]/(f_1 + yg_1, \dotsc, f_m + yg_m). $$

Both $B_{g_1, \dotsc, g_m}$ and $B'_{g_1, \dotsc, g_m}$ have $A$ as a quotient obtained by killing the auxiliary variables $y_1, \dotsc, y_m$ or $y$. My questions are:

1. Can one always choose $g_1, \dotsc, g_m$ in such a way that $\mathrm{Spec}(B_{g_1, \dotsc, g_m}) \setminus \mathrm{Spec}(A)$ be Cohen-Macaulay (or, even better, $B_{g_1, \dotsc, g_m}$ be Cohen-Macaulay)?
2. Can one always choose $g_1, \dotsc, g_m$ in such a way that $\mathrm{Spec}(B'_{g_1, \dotsc, g_m}) \setminus \mathrm{Spec}(A)$ be Cohen-Macaulay (or, even better, $B'_{g_1, \dotsc, g_m}$ be Cohen-Macaulay)?

This is somewhat a follow up to my previous question Embedding a given affine variety as a divisor where we found out that it is not in general possible to do the above by also requiring that, say, $y$ be a nonzerodivisor.