As stated, I believe the answer is no. Set $q_1=x_1^2$, $q_2=x_2^4$, and consider $I=(q_1,q_2)$. Let $p_1=q_1$, and let $p_2=x_1p_1$. Then $p_1$ is (up to a constant) the only degree 2 polynomial in $I$, and $p_2$ is (up to a linear combination of multiples of $p_1$) the only polynomial of degree 3 in $I$. But they do not generate $I$.

If we add the requirement that $p_2$ is not a multiple of $p_1$, as Dave Witte Morris suggests, we can use the following example. Let $q_1=x_1x_2$ and $q_2=x_2^4+x_1^2$. Let $p_1=x_1x_2$ and $p_2=x_1^3=x_1q_2-x_2^3q_1$. The ideal $I=(q_1,q_2)$ is not generated by $p_1$ and $p_2$ since $q_2$ not divisible by $x_1$. Using normal forms, we can check directly that $x_1x_2=0$, $x_1^3=0$, $x_2^4=-x_1^2$ is a reduction system for the ring $\mathbb{R}[x_1,x_2,x_3,x_4,x_5]/I$, and so there are no more polynomials in degrees 2 or 3 in $I$ than those given by linear combinations of multiples of $p_1$ and $p_2$.