No. Take $n=2$ and define $$f(t,x) = \frac{1}{\sqrt{t^2 + \left| x \right|^2}}$$ Then for $1 \leq q < 2$ \begin{align*} \int_{B_1(0)} f(t,x)^q dx &= 2\pi \int_0^1 \frac{r}{(t^2 + r^2)^{q/2}} dr \\ &= 2\pi \left. \frac{1}{q+2} (t^2 + r^2)^{-q/2 + 1} \right|_0^1 \end{align*}

and so $f(t,x) \in L^\infty(0,T;L^2)$

If $p=2$ then \begin{align*} \int_{B_1(0)} f(t,x)^2 &= 2\pi \left[\log(t^2 + 1) - \log(t^2)\right] \end{align*}

Now, $\log(t)$ is integrable on $(0,T)$ and so $f(t,x) \in L^2(Q)$. However, $\log(t)$ is clearly not bounded and so you do $\bf{not}$ get $f(t,x) \in L^\infty(0,T;L^2)$ as you wanted.

No. Take $n=2$ and define $$f(t,x) = \frac{1}{\sqrt{t^2 + \left| x \right|^2}}$$ Then for $1 \leq q < 2$ \begin{align*} \int_{B_1(0)} f(t,x)^q dx &= 2\pi \int_0^1 \frac{r}{(t^2 + r^2)^{q/2}} dr \\ &= 2\pi \left. \frac{1}{q+2} (t^2 + r^2)^{-q/2 + 1} \right|_0^1 \end{align*}

and so $f(t,x) \in L^\infty(0,T;L^q)$

If $p=2$ then \begin{align*} \int_{B_1(0)} f(t,x)^2 &= 2\pi \left[\log(t^2 + 1) - \log(t^2)\right] \end{align*}

Now, $\log(t)$ is integrable on $(0,T)$ and so $f(t,x) \in L^2(Q)$. However, $\log(t)$ is clearly not bounded and so you do $\bf{not}$ get $f(t,x) \in L^\infty(0,T;L^2)$ as you wanted.