The minimum (which is an infimum) is $-\infty$.

We have that $\limsup_{t\to 0} B_t/\sqrt{t}=\infty$ (this follows in particular from the LIL). This means that there exists $t_M\in (0,1)$ arbitrarily small so that $B_{t_M}>M \sqrt{t_K}$. Let us disregard for a moment that $t_M$ is not a stopping time.

Take now $f(t)=(1/\sqrt{t_M}) 1_{t<t_M}$. Then $\int f(t)^2 dt=1$, while $\int f(t) dB_t= M$. Then $J(f)=\int f(t) dt-2\int f(t) dB_t=-M+1\to_{M\to\infty} -\infty$.

Now, in that construction $t_M$ is not a stopping time, so the definition of the stochastic integral requires some care. However, we can modify and simplify the construction as follows. For $K$ the square root of an integer, consider the two functions $f_{1,K}=K1_{t<1/K^2}$ and $f_{2,K}=-f_{1,K}$. The value of $\min(J(f_{1,K}),J(f_{2,K}))=1-|B_{1/K^2} K|$. Then, again by an application of the LIL on the discrete sequence $j$ (using that $t B_{1/t^2}$ is a Brownian motion), one has $$ \inf_{i=1,2, j\in N} J(f_{i,\sqrt{j}})=-\infty,$$ almost surely.

The minimum (which is an infimum) is $-\infty$.

We have that $\limsup_{t\to 0} B_t/\sqrt{t}=\infty$ (this follows in particular from the LIL). This means that there exists $t_M\in (0,1)$ arbitrarily small so that $B_{t_M}>M \sqrt{t_K}$. Let us disregard for a moment that $t_M$ is not a stopping time.

Take now $f(t)=(1/\sqrt{t_M}) 1_{t<t_M}$. Then $\int f(t)^2 dt=1$, while $\int f(t) dB_t= M$. Then $J(f)=\int f(t)^2 dt-2\int f(t) dB_t=-2 M+1\to_{M\to\infty} -\infty$.

Now, in that construction $t_M$ is not a stopping time, so the definition of the stochastic integral requires some care. However, we can modify and simplify the construction as follows. For $K$ the square root of an integer, consider the two functions $f_{1,K}=K1_{t<1/K^2}$ and $f_{2,K}=-f_{1,K}$. The value of $\min(J(f_{1,K}),J(f_{2,K}))=1-|B_{1/K^2} K|$. Then, again by an application of the LIL on the discrete sequence $j$ (using that $t B_{1/t^2}$ is a Brownian motion), one has $$ \inf_{i=1,2, j\in N} J(f_{i,\sqrt{j}})=-\infty,$$ almost surely.