Properties of heat equation

The answer is no. A counterexample has the form $$u(x,t)=\sum_0^\infty \frac{g^{(n)}(t)}{((2n+1)!)^2}(r-R)^{2n+1},$$ where $r>R$ is the distance from the origin in $R^2$. First one shows that this function satisfies the heat equation formally. Second, there exists an infinitely differentiable $g\not\equiv 0$ such that $g^{(n)}(0)=0$ for all $n\geq 0$, and such that the above series converges. So our function is not zero, satisfies the heat equation in the exterior of the ball, and has zero boundary and initial conditions. For the existence of such $g$ one can refer to http://www-bcf.usc.edu/~lototsky/MATH445/NonUniqueness-HeatEq.pdf

WLOG, $u$ is negavive somewhere, otherwise replace it by $-u$. Then add a small positive constant to $u$, the boundary and initial conditions will be positive, but $u$ is still negative somewhere.

As a companion to Alexandre Eremenko's answer, the answer to your question is "yes" if we assume in addition that $u$ satisfies the growth condition $|u| < Ae^{M|x|^2}$ for some $A,\,M > 0$.

To see this assume for simplicity that $a = 0$ and $T = 1$. Let $h_R = -t^{-1/2}e^{-\frac{(x-R)^2}{4t}}$ and let $b_R = Ae^{MR^2}h_R$. By the comparison principle on bounded domains we have $u \geq b_R$ on $[0,\,R] \times [0,\,1]$. A short computation shows that $b_R \geq -8A\sqrt{M}e^{-3MR^2}$ on $[0,\,R/2] \times [0,\, (64M)^{-1}]$. Taking $R \rightarrow \infty$ we conclude that $u \geq 0$ for $t \in [0,\,(64M)^{-1}]$, and repeating the argument finitely many times we get $u \geq 0$ on $[0,\,T]$. That $u > 0$ follows from the strong maximum principle.