Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $$P(-1)=-a+b-c+d=-9.$$ Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.
We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or $$a'+b+c'+d=9.$$ By the stars and bars argument, there are $\binom{9+4-1}{4-1}=\boxed{\textbf{(D) } 220}$ ordered quadruples $(a',b,c',d).$
~pieater314159 ~MRENTHUSIASM

1. Let $P(x) = ax^3 + bx^2 + cx + d$. We are given that $P(-1) = -9$.
2. Substituting $x = -1$ into $P(x)$, we get:
$$P(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d$$
3. We need $-a + b - c + d = -9$. Rearranging, we get:
$$b + d = a + c + 9$$
4. Let $s = a + c + 9$. Then $b + d = s$. The possible values of $s$ range from 9 to 18 (since $a, b, c, d$ are digits from 0 to 9).
5. For each fixed $s$, we need to count the number of valid pairs $(a, c)$ and $(b, d)$ such that:
$$a + c = s - 9 \quad \text{and} \quad b + d = s$$
6. The number of pairs $(a, c)$ such that $a + c = s - 9$ is $s - 8$ (since $a$ and $c$ are digits from 0 to 9).
7. The number of pairs $(b, d)$ such that $b + d = s$ is $19 - s$ (since $b$ and $d$ are digits from 0 to 9).
8. Therefore, for each $s$, the number of valid polynomials is:
$$(s - 8)(19 - s)$$
9. We need to sum this product over all possible values of $s$ from 9 to 18:
$$\sum_{s=9}^{18} (s - 8)(19 - s)$$
10. Simplifying the sum:
$$\sum_{s=9}^{18} (s - 8)(19 - s) = \sum_{s=0}^{9} (s + 1)(10 - s)$$
11. Calculating each term:
$$\begin{aligned} & (0 + 1)(10 - 0) = 1 \cdot 10 = 10 \\ & (1 + 1)(10 - 1) = 2 \cdot 9 = 18 \\ & (2 + 1)(10 - 2) = 3 \cdot 8 = 24 \\ & (3 + 1)(10 - 3) = 4 \cdot 7 = 28 \\ & (4 + 1)(10 - 4) = 5 \cdot 6 = 30 \\ & (5 + 1)(10 - 5) = 6 \cdot 5 = 30 \\ & (6 + 1)(10 - 6) = 7 \cdot 4 = 28 \\ & (7 + 1)(10 - 7) = 8 \cdot 3 = 24 \\ & (8 + 1)(10 - 8) = 9 \cdot 2 = 18 \\ & (9 + 1)(10 - 9) = 10 \cdot 1 = 10 \\ \end{aligned}$$
12. Summing these values:
$$10 + 18 + 24 + 28 + 30 + 30 + 28 + 24 + 18 + 10 = 220$$

The final answer is $\boxed{220}$