Is there a prime every year if YYYYMMDD is treated as a base-$10$ number?

It is false, as you can construct sequences of composite numbers of arbitrary length $n$, e.g., $$(n+1)! + 2, \ldots, (n + 1)! + (n + 1) :$$ Pick such a sequence of length at least $19999$, so that it contains a subsequence of consecutive numbers ending in $0000, \ldots, 9999$. Then, the first number in that subsequence is $10\,000 \cdot y$ for some integer $y$, and all the numbers given by date strings from the year $y$ are composite.

Your conjecture is false.

Let $S=\left\{s_1,s_2,\ldots,s_{366}\right\}$ be the set of all numbers of the form MMDD, including $229$. We can restate your problem in the following marginally weaker form (it's weaker since not all years are leap years):

For every $k$, at least one of the numbers $$10000k+s$$ for $s\in S$ is a prime number.

However, this can easily be proven false. If we take $p_1,p_2,\ldots,p_{366}$ to be $366$ distinct prime numbers, different from $2$ and $5$, by the Chinese Remainder Theorem, there exists a $Y$ such that $$Y\equiv-s_1\cdot10000^{-1}\pmod{p_1}$$ $$Y\equiv-s_2\cdot10000^{-1}\pmod{p_2}$$ $$\vdots$$ $$Y\equiv-s_{366}\cdot10000^{-1}\pmod{p_{366}}$$ For this year $Y$, every single number $10000Y+s_i$ that we can generate will be divisible by the corresponding $p_i$ (while at the same time being much bigger than it), meaning that this year will have no primes.

Here's an alternate proof, much more technical, but much more faster. If every year had a prime, then $\pi(n)\in O(n)$, which is false by the Prime Number Theorem.