A cleaner way to handle calendar-shift problems is to assign each day of the week a number and work with arithmetic modulo 7 instead of counting day by day.
Let Sunday $=0$, Monday $=1$, Tuesday $=2$, Wednesday $=3$, Thursday $=4$, Friday $=5$, Saturday $=6$. We are told June 2 corresponds to Thursday, so June 2 $\equiv 4 \pmod 7$.
The number of days from June 2 to July 3 is $(30-2)+3 = 31$ days, since June has 30 days.
The day-of-week value for July 3 is then $(4+31) \bmod 7$. Since $31 = 28+3$ and $28$ is exactly $4 \times 7$, this reduces to $(4+3) \bmod 7 = 7 \bmod 7 = 0$.
A value of $0$ corresponds to Sunday in the numbering scheme set up above.
\[\boxed{\text{July 3 is a Sunday}}\]