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

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.

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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.

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.