jmc

We begin by converting $403_{10}$ into base-7. Since $7^3=343$ is the largest power of 7 that is less than 403, and it can go into the given number once, the coefficient of the $7^3$ term will be 1. From here, we are left with a remainder of $403-343=60$. The largest power of 7 less than this number is $7^2=49$, and the largest multiple of 49 that is less than 60 is $1\cdot 49=49$ itself. This leaves us with $60-49=11$, which we can express as $1\cdot 7^1+4\cdot 7^0$. So, we find that $403_{10}=1\cdot 7^3+1\cdot 7^2+1\cdot 7^1+4\cdot 7^0=1114_7$, which only has $\boxed{1}$ even digit.