No, it is not.
A downhill number can always be written as $a-b_{1}-b_{2}-\ldots-b_{9}$, where $a$ is of the form $99 \ldots 99$ and each $b_{i}$ either equals 0 or is of the form $\overline{11 \ldots 11}$.

Let $n$ be a positive integer. The numbers of the form $99 \ldots 99$ yield at most $n$ different remainders upon division by $2^{n}$, as do the numbers of the form $11 \ldots 11$. Therefore, downhill numbers yield at most $n(n+1)^{9}$ different remainders upon division by $2^{n}$.

Let $n$ be so large that $n(n+1)^{9}<2^{n}$. ( $n=63$ works: $63 \times 64^{9}<64^{10}=2^{60}<2^{63}$.) Let $0 \leq r<2^{n}$ be such that no downhill number is congruent to $r$ modulo $2^{n}$.

Consider the polynomial

$$P(x)=\frac{1}{2 \times\left(2^{n}-1\right)!} \prod_{1 \leq i<2^{n}}(x-r+i) .$$

We have that $P(r)=\frac{1}{2}$ is not an integer.
Let, then, $x$ be a downhill number. The number $(x-r+1) \ldots\left(x-r+2^{n}-1\right)$ is a multiple of $\left(2^{n}-1\right)$ ! (as a product of $2^{n}-1$ consecutive integers); therefore, $2 P(x)$ is an integer. On the other hand, the number $(x-r)(x-r+1) \ldots\left(x-r+2^{n}-1\right)$ is a multiple of $2^{n}$ ! (as a product of $2^{n}$ consecutive integers); therefore, $2(x-r) P(x)$ is an integer multiple of $2^{n}$. Since $x$ is downhill, $x-r$ is not divisible by $2^{n}$. Therefore, $2 P(x)$ is even and $P(x)$ is an integer.

Alternative version. A positive integer $n$ is uphill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \leq a_{k-1} \leq \ldots \leq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and uphill-integer-valued if $P(n)$ is an integer for all uphill positive integers $n$. Is it true that every uphill-integer-valued polynomial is also integer-valued?