To expand on the answer in my comment, the proportion of integers $a$ for which $4a+1,4a+2,4a+3$ are all squarefree is $\prod\_{p\not=2}(1-3/p^2)$, with the product taken over all odd primes $p$. As this product converges to a positive limit, there are infinitely many such $a$. A quick heuristic is to look modulo $p^2$. For odd prime $p$, precisely $p^2-3$ of the possible $p^2$ values of $a$ mod $p^2$ lead to $4a+1,4a+2,4a+3$ all being nonzero mod $p^2$, so this has probability $1-3/p^2$. Independence of mod $p^2$ arithmetic as $p$ runs through the primes suggests the claimed limit.
More precisely, if $\phi(n)$ is the number of positive integers $a\le n$ with $4a+1,4a+2,4a+3$ squarefree then $$ \frac{\phi(n)}{n}\to\prod\_{p\not=2}(1-3/p^2).\qquad\qquad{\rm(1)} $$ It's not too hard to turn this heuristic into a rigorous argument. If we let $\phi_N(n)$ denote the number of $a\le n$ such that none of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for a prime $p < N$, then the Chinese remainder theorem says that we get equality $$ \frac{\phi_N(n)}{n}=\prod\_{\substack{p\not=2,\\\\ p < N}}(1-3/p^2).\qquad\qquad{\rm(2)} $$ wherever $n$ is a multiple of $\prod\_{\substack{p\not=2,\\\\ p < N}}p^2$ and, therefore, the error in (2) is of order $1/n$ for arbitrary $n$. It only needs to be shown that ignoring primes $p\ge N$ leads to an error which is vanishingly small as $N$ is made large. In fact, the number of $a \le n$ which are a multiple of $p^2$ is $\left\lfloor\frac{n}{p^2}\right\rfloor\le \frac{n}{p^2}$. The number of $a\le n$ which is a multiple of $p^2$ for some prime $p\ge N$ is bounded by $n\sum\_{p\ge N}p^{-2}$. So, the proportion of $a\le n$ for which one of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for an odd prime $p\ge N$ is bounded by $\frac{4n+3}{n}\sum_{p\ge N}p^{-2}\sim (4+3/n)/(N\log N)$. This means that $\phi_N(n)/n\to\phi(n)/n$ uniformly in $n$ as $N\to\infty$, and the limit (1) follows from approximating by $\phi_N$.