The appearance of the Heaviside $\theta$ is due to imposing a condition of convergence on the circular arc section of the contour, $T_R$ as you call it. When $t \gt 0$, substitute $p = R e^{i \phi}$ and we have
$$\int_{T_R} dp \frac{e^{p t}}{p} = i R \int_{\pi/2}^{3 \pi/2} d\phi\, e^{i \phi} \frac{e^{R t \cos{\phi}} e^{i R t \sin{\phi}}}{i R e^{i \phi}}$$
Note that $\cos{\phi} \lt 0$ when $\phi\in (\pi/2, 3 \pi/2)$, so that the integral vanishes in the limit of $R \to \infty$ by Jordan's lemma only when $t \gt 0$. In this region, there is a pole at $p=0$ and by the residue theorem, $f(t) = 1$ here.
When $t \lt 0$, however, the integral about $T_R$ does not vanish, but blows up. Thus, we may not use this contour. Rather, we go the other way, using a new segment $S_R$ which goes to the right of $\Re{z} = \sigma$. Here, $\cos{\phi} \gt 0$ and so Jordan's lemma applies when $t \lt 0$ (and not when $t \gt 0$). In this region, there are no poles (this is why we always choose $\sigma$ to be to the right of all of the poles of $\hat{f}(p)$), so the integral over the closed contour is zero.
Therefore, the ILT of $\hat{f}(p) = 1/p$ is $f(t) = \theta(t)$.
The way to attack a problem like this is via Cauchy's theorem on a properly distorted Bromwich contour. Here, we want our contour to avoid the branch point at $z=0$. This, we consider
$$\oint_C dz \, z^{-a-1} e^{-z^a} e^{z t} $$
where $a \in (0,1)$ and $C$ is the following contour:
We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis. There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.
$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.
$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.
$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.
$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.
$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.
$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.
When $t \gt 0$, the integral over the contours $C_2$ and $C_6$ vanish in the limit as $R \to \infty$.
The contour integral is thus equal to, in this limit,
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} + e^{-i \pi a} \int_{\infty}^{\epsilon} dx \, x^{-a-1} e^{-e^{i \pi a} x^a} e^{-x t} \\ + i \epsilon^{-a} \int_{\pi}^{-\pi} d\phi \, e^{-i a \phi} e^{-\epsilon^a e^{i a \phi}} e^{\epsilon t e^{i \phi}} + e^{i a \pi} \int_{\epsilon}^{\infty} dx \, x^{-a-1} e^{-e^{-i \pi a} x^a} e^{-x t}$$
Note that there is an apparent singularity at $\epsilon = 0$; however, the divergences cancel in the limit as $\epsilon \to 0$.
In this limit, the third integral has the following leading behavior:
$$-i \frac{2}{a} \epsilon^{-a} \sin{\pi a} +i 2 \pi$$
Rescaling and combining the second and fourth integrals, we get for the contour integral:
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} -i 2 t^a \operatorname{Im}{\left [e^{-i \pi a}\int_{\epsilon t}^{\infty} \frac{du}{u^{1+a}} e^{-u} e^{-e^{i \pi a} t^{-a} u^a} \right]}-i \frac{2}{a} \epsilon^{-a} \sin{\pi a} +i 2 \pi $$
We may Taylor expand the second exponential in the integrand because it is subdominant to the first exponential (at least for the first $n$ terms, where $n$ is that largest integer such that $\lfloor n a \rfloor = 0$). We need only expand to the first two terms to treat the limit as $\epsilon \to 0$. Note that
$$\int_{\epsilon t}^{\infty} \frac{du}{u^{1+a}} e^{-u} = \frac{t^{-a}}{a} \epsilon^{-a} + \Gamma(-a) + O \left ( \epsilon^a \right ) $$
The second term produces
$$-e^{i \pi a} t^{-a}\int_{\epsilon t}^{\infty} \frac{du}{u} e^{-u} $$
Because the exponentials outside the integral cancel, the imaginary part of the term is zero. Thus, we now take the limit as $\epsilon \to 0$. Because the contour integral is zero by Cauchy's theorem, we get for the ILT,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} =
\frac{t^a}{\pi} \operatorname{Im}{\left [e^{-i \pi a}\int_{0}^{\infty} \frac{du}{u^{1+a}} e^{-u} \left ( e^{-e^{i \pi a} t^{-a} u^a} - 1 + e^{i \pi a} t^{-a} u^a \right ) \right]} + \frac{t^a}{\Gamma(1+a)} - 1$$
ADDENDUM
Even though the above result is suitable for numerical calculation, we can illustrate the above result with an analytical example. Consider the case $a=1/2$. Subbing $u=x^2$ and taking the imaginary part of the integral, we end up with
$$\operatorname{Im}{\left [e^{-i \pi a}\int_{0}^{\infty} \frac{du}{u^{1+a}} e^{-u} \left ( e^{-e^{i \pi a} t^{-a} u^a} - 1 + e^{i \pi a} t^{-a} u^a \right ) \right]} = 2 \int_{-\infty}^{\infty} dx \, e^{-x^2} \frac{\sin^2{\beta x}}{x^2}$$
where $\beta = 1/(2 \sqrt{t})$.
The latter integral may be evaluated using Parseval's theorem, because the individual factors of the integrand are inverse Fourier transforms of simple functions. For example,
$$\int_{-\infty}^{\infty} dx \, e^{-x^2} e^{i k x} = \sqrt{\pi} e^{-k^2/4} $$
$$\int_{-\infty}^{\infty} dx \, \frac{\sin^2{\beta x}}{x^2} e^{i k x} =\begin{cases} \pi \beta \left ( 1-\frac{|k|}{2 \beta} \right ) & |k| \lt 2 \beta \\ 0 & |k| \gt 2 \beta \end{cases}$$
The integral is then equal to
$$2 \frac1{2 \pi} \sqrt{\pi} \pi \beta \int_{-2 \beta}^{2 \beta} dk \, \left ( 1-\frac{|k|}{2 \beta} \right ) e^{-k^2/4} = \sqrt{\pi} \beta \int_0^{2 \beta} dk \, \left ( 1-\frac{k}{2 \beta} \right ) e^{-k^2/4}$$
The evaluation is fairly straightforward using the definition of the error function. The result is, for the integral,
$$2 \pi \beta \operatorname{erf}{\beta} - 2 \sqrt{\pi} \left (1-e^{-\beta^2}\right ) $$
Now plugging this back into the main result above and using $\beta = 1/(2 \sqrt{t})$, we get that
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{-s^{1/2}} e^{s t} &= \operatorname{erf}{\left ( \frac1{2 \sqrt{t}} \right )} + \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - 1 \\ &= \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - \operatorname{erfc}{\left ( \frac1{2 \sqrt{t}} \right )}\end{align}$$
Best Answer
The inverse Laplace transform of $F(s)=\frac{s}{s^2+a^2}$ is given by
$$f(x)=\frac1{2\pi i}\int_{\alpha-i\infty}^{\alpha+i\infty}F(s)e^{sx}ds$$ where $\alpha=\Re (s)$. The integral is carried out by contour integration.
The contour from $\alpha-i\infty$ to $\alpha+i\infty$ is referred to as the Bromwich contour, and $\alpha$ is taken to the right of all singularities in order to insure
$$\int_0^\infty e^{-\alpha x}\vert f(x)\vert dx<\infty$$ Hence, using the residue theorem and as $R\to\infty$ and so $T\to\infty$, we get $$f(x)=\sum_{j=1}^2 \operatorname{Res}\left(\frac{se^{sx}}{s^2+a^2},s_j\right),\quad s_1=ia,\quad s_2=-ia$$ so $$f(x)=\frac{ia e^{iax}}{2ia}+\frac{iae^{-iax}}{2ia}=cos(ax)\quad (x>0)$$